Approximation and interpolation of deep neural networks
In this paper, we prove that in the overparametrized regime, deep neural
network provide universal approximations and can interpolate any data set, as
long as the activation function is locally in $L^1(\RR)$ and not an affine
function.
  Additionally, if the activation function is smooth and such an interpolation
networks exists, then the set of parameters which interpolate forms a manifold.
Furthermore, we give a characterization of the Hessian of the loss function
evaluated at the interpolation points.
  In the last section, we provide a practical probabilistic method of finding
such a point under general conditions on the activation function.
Vlad-Raul Constantinescu, Ionel Popescu
# Interpolation property of shallow neural networks

###### Abstract

We study the geometry of global minima of the loss landscape of overparametrized neural networks. In most optimization problems, the loss function is convex, in which case we only have a global minima, or nonconvex, with a discrete number of global minima. In this paper, we prove that in the overparametrized regime, a shallow neural network can interpolate any data set, i.e. the loss function has a global minimum value equal to zero as long as the activation function is not a polynomial of small degree. Additionally, if such a global minimum exists, then the locus of global minima has infinitely many points. Furthermore, we give a characterization of the Hessian of the loss function evaluated at the global minima, and in the last section, we provide a practical probabilistic method of finding the interpolation point.

## 1 Introduction

We study the geometry of global minima of the loss landscape of overparametrized neural networks. In the light of the interpolation threshold outlined in [1] one of the important issues in neural networks is to have guarantees that the interpolation is indeed achieved. We tackle this problem for the case of shallow neural network and show that this holds true in general as long as the activation is not a polynomial of low degree.

Standard optimization problems are done in the case the loss function is convex in which case we only have a global minima. Another class of optimization problems is for nonconvex loss functions which has a discrete number of global minima.

Recently there has been interesting progress aimed at understanding the locus of the global minima for overparametrized neural networks ([3], [6]) when the activation function is continuous. In this paper, we generalize these results in section 2 for a larger class of activation functions. More precisely, we prove that in the overparametrized regime, we can interpolate any data set consisting of \(d\) points with a shallow neural network having at least \(d\) neurons on the hidden layer and with an activation function which is locally integrable and not almost everywhere a polynomial of degree at most \(d-2\). In addition, if the activation function is also smooth, the locus of global minima of the loss landscape of an over-parametrized neural network is a submanifold of \(\mathbb{R}^{n}\). More precisely, if a neural net has \(n\) parameters and is trained on \(d\) data points, where \(n>d\), then the locus \(M\) of global minima is an \(n-d\) dimensional submanifold of \(\mathbb{R}^{n}\). If the activation function \(\sigma\) is a non-constant polynomial, the interpolation problem depends very much on the data set and the degree of \(\sigma\). In section 3 we provide similar results for this class of activation functions. Additionally, we give a description in section 4 for the eigenspectrum of the Hessian of the loss function of a neural net evaluated at the global minima.

The next question that arises is how can one find such an interpolation point? One of the numerical methods that is used for training neural networks is the (stochastic) gradient descent method. For the convex case, there is a large literature on convergence results for (stochastic) gradient descent, see for example [2]. However, in the non-convex scenario, first-order methods like gradient descent can get stuck at a saddle point. In section 5 we give an answer to this question by reducing the minimization of a non-convex loss function to a simple linear regression problem. In [5], this is done for shallow neural networks, using a smooth activation function with a bounded derivative and a number of hidden nodes of order \(O(d\log^{2}(d))\). Using similar techniques as in [5], we extend this result to shallow neural networks with a continuous activation function that is not a polynomial of degree less than \(d-2\). The reduction is done by a random initialization of the input-to-hidden weights, and optimizing over the output layer. Our result requires that the number of hidden is of order \(O(d\log(d))\).
## 2 Interpolation with non-polynomial activation function

We consider a neural network of any architecture (eg., feedforward, convolutional, etc.), with weights \(w=(w_{1},w_{2},\ldots)\) and biases \(b=(b_{1},b_{2},\ldots)\). The number of weights and biases is \(n\), and we train our neural network on \(d\) data points \((x_{i},y_{i})_{i=\overline{1,d}}\), where \(x_{i}\in\mathbb{R}^{p}\) and \(y_{i}\in\mathbb{R}\). We assume that the \(x_{i}\) are distinct and our neural network is overparametrized, i.e. \(n>d\).

We denote by \(f_{w,b}\) the function given by our neural network. For each data point \((x_{i},y_{i})\), we define \(f_{i}(w,b)=f_{w,b}(x_{i})-y_{i}\). We suppose that each \(f_{i}(w,b)\) is smooth in \(w\) and \(b\). For example, if our neural network is feedforward, the smoothness of the activation function \(\sigma\) implies the smoothness of \(f_{i}(w,b)\).

For the training of our neural net, we use the mean squared loss function

\[L(w,b)=\sum_{i=1}^{d}f_{i}(w,b)^{2}\]

From our definition of the loss function, we observe that \(L(w,b)\geq 0\). If \(M=L^{-1}(0)\) is nonempty, then \(M\) is the locus of global minima of \(L\). Also, the locus of global minima can be written as

\[M=\bigcap_{i=1}^{d}M_{i},\]

where

\[M_{i}=f_{i}^{-1}(0)\]

The following theorem is a result of [3] which we state here for the case of smooth activation functions.

**Theorem 2.1**.: _In the framework above, the set \(M=L^{-1}(0)\) is generically (that is, possibly after an arbitrarily small change to the data set) a smooth \(n-d\) dimensional submanifold (possibly empty) of \(\mathbb{R}^{n}\)._

In this paper, we will prove that for a class of feedforward neural networks, the set \(M=L^{-1}(0)\) is non-empty. In this context, \(f_{w,b}\) is written in matrix form as

\[f_{w,b}(x)=W_{l}\sigma(W_{l-1}\sigma(\ldots\sigma(W_{1}x-b_{1})\ldots)-b_{l-1} )-b_{l},\]

where \(W_{i}\in\mathcal{M}_{n_{i-1}\times n_{i}}(\mathbb{R}),b_{i}\in\mathbb{R}^{n_{i}}\) and \(n_{0}=p,n_{l}=1\). Moreover, we use the convention that \(\sigma\) applied to a vector is simply the component-wise evaluation:

\[\sigma(v_{1},v_{2},\ldots,v_{k})=(\sigma(v_{1}),\sigma(v_{2}),\ldots,\sigma(v _{k})).\]

When the activation function \(\sigma\) is continuous and not a polynomial, any shallow neural network, i.e. a feedforward neural network with one hidden layer, can interpolate any data set [6]. In this paper, we will prove the same result for activation functions \(\sigma\), which satisfy the following assumption:

**Assumption 2.2**.: _The activation function \(\sigma\) is locally integrable, i.e \(\sigma\in L^{1}_{loc}(\mathbb{R})\), and is almost surely not a polynomial of degree less or equal than \(d-2\), i.e. there exists no polynomial \(P\) of degree at most \(d-2\) such that \(\sigma=P\) almost surely._

**Proposition 2.3**.: _Let \((x_{i},y_{i})_{i=\overline{1,d}}\) be a data set with \(x_{i}\in\mathbb{R}^{p},y_{i}\in\mathbb{R}\), and with \(x_{i}\) assumed distinct. Assume that \(\sigma\) satisfies Assumption 2.2. Then, for any \(h\geq d\), there exists a shallow neural network with width \(h\geq d\), with activation function \(\sigma\) such that it interpolates our data set, i.e. \(f_{w,b}(x_{i})=y_{i}\) for all \(i\)._

Proof.: The idea is to refine the proof of Theorem 5.1 from [6]. The output function of a shallow neural network is written in the matrix form as

\[f_{w,b}(x)=v^{T}\sigma(Wx-b)-b^{\prime},\]

where \(W\in\mathcal{M}_{h\times p}(\mathbb{R}),v\in\mathbb{R}^{h},b\in\mathbb{R}^{h}\) and \(b^{\prime}\in\mathbb{R}\). The entries of \(w\) and \(v\) are the weights, and the entries of \(b,b^{\prime}\) are the biases. For our neural net, we take \(b^{\prime}=0\) and \(h=d\). If \(h>d\), then we set the weights and biases after the first \(d\) nodes to be equal to zero. Hence, we can reduce our construction to \(h=d\). Let \(w_{1},\ldots,w_{d}\) be the lines of \(W\) and \(v=(v_{1},\ldots,v_{d})\). Since the \(x_{i}\) are distinct, we can find a vector \(w\in\mathbb{R}^{p}\) such that \(w^{T}x_{i}=t_{i}\) are distinct for all \(i\). We set \(w_{i}=a_{i}^{T}w\) for some \(a_{i}\in\mathbb{R}\), \(i=\overline{1,d}\). Therefore, we have to show that there exists \((a_{i},b_{i},v_{i})_{i=\overline{1,d}}\) such that

\[\sum_{j=1}^{d}v_{j}\sigma(a_{j}t_{i}-b_{j})=y_{i},\]
for all \(i\). This interpolation problem is equivalent to proving the linear independence (over \(a\) and \(b\)) of the \(d\) functions \(\sigma(at_{i}-b)\). If we have linear independence of these functions, then we can find \((a_{i},b_{i})_{i=\overline{1,d}}\) such that the matrix system of our interpolation problem

\[\left(\begin{array}{ccc}\sigma(a_{1}t_{1}-b_{1})&\ldots&\sigma(a_{d}t_{1}-b_{ d})\\ \vdots&\ddots&\vdots\\ \sigma(a_{1}t_{d}-b_{1})&\ldots&\sigma(a_{d}t_{d}-b_{d})\end{array}\right)\]

is nonsingular. And from here we can determine \((v_{1},\ldots,v_{d})\) uniquely. Suppose that our \(d\) functions are linearly dependent. This means that we can find nontrivial coefficients \((c_{i})_{i=\overline{1,d}}\) such that

\[\sum_{i=1}^{d}c_{i}\sigma(at_{i}-b)=0. \tag{1}\]

Let \(\zeta\in C_{\circ}^{\infty}(\mathbb{R},[0,\infty))\), i.e. \(\zeta\) is non-negative, infinitely differentiable with compact support and \(\int_{\mathbb{R}}\zeta(x)dx=1\). We define for \(\epsilon>0\), the following function

\[\sigma_{\epsilon}(t)=\int_{\mathbb{R}}\frac{1}{\epsilon}\zeta\left(\frac{t-x }{\epsilon}\right)\sigma(x)dx\]

Since \(\sigma\in L^{1}_{loc}(\mathbb{R})\), standard arguments show that

\[\sigma_{\epsilon}\xrightarrow[\epsilon\to 0]{1}\sigma\]

In particular, we also have along a subsequence \(\epsilon_{n}\) which converges to \(0\) such that \(\sigma_{\epsilon_{n}}\) converges to \(\sigma\) almost surely. The key observation is that if \(\sigma_{\epsilon_{n}}\) is a polynomial of degree less than \(d-2\) for every \(n\), then in the limit we also have that \(\sigma\) is almost surely a polynomial of degree at most \(d-2\).

Consequently, we can reduce the problem to the case where \(\sigma\) is replaced by \(\sigma_{\epsilon}\) for some \(\epsilon>0\). Using now (1) we will continue to have the same relation also for \(\sigma_{\epsilon}\). Thus from now on we simply assume that \(\sigma\) is smooth and (1) is satisfied. If we differentiate \(k\) times relation (1) with respect to \(a\), we get

\[\sum_{i=1}^{d}c_{i}t_{i}^{k}\sigma^{(k)}(at_{i}-b)=0.\]

Since \(\sigma\) is not a polynomial of degree less or equal than \(d-2\), for any \(k=\overline{0,d-1}\) we can find \(b_{k}\in\mathbb{R}\) such that \(\sigma^{(k)}(-b_{k})\neq 0\). Taking \(a=0\) and \(b=b_{k}\) for each equation, we get a system of \(d\) equations

\[\sum_{i=1}^{d}c_{i}t_{i}^{k}=0, \tag{2}\]

for each \(k=\overline{0,d-1}\). Since the matrix system of (2) is a Vandermonde matrix, and the \(t_{i}\) are distinct, we get that all \(c_{i}\) must be equal to \(0\), which is a contradiction. 

Since a shallow neural network can be embedded in a deeper feedforward neural network, we have the following Corrolary of Proposition 2.3.

**Corollary 2.4**.: _Let \((x_{i},y_{i})_{i=\overline{1,d}}\) be a data set with \(x_{i}\in\mathbb{R}^{p},y_{i}\in\mathbb{R}\), and the \(x_{i}\) are distinct. Assume that the activation function \(\sigma\) satisfies Assumption 2.2 and it is one-to-one on some interval \(I\) of the real line. Then, in the family of feedforward neural nets of \(l\) hidden layers, with the last hidden layer having \(h\geq d\) neurons, and remaining hidden layers of any width, we can find one that interpolates our data set, i.e. \(f_{w,b}(x_{i})=y_{i}\) for all \(i\)._

Proof.: By the hypothesis, \(\sigma\) is injective on \(I\). The idea is to first the map the input into the \(l-2\) hidden layer one-to-one and then consider these as the input data for a shallow neural network. To this end, let \(f_{k}\) denote the output function of the first \(k\) hidden layers before the activation is applied, i.e.

\[f_{k}(x)=W_{k}\sigma(W_{k-1}\sigma(\ldots\sigma(W_{1}x-b_{1})\ldots)-b_{k-1})- b_{k}.\]

Notice that \(f_{1}(x)=W_{1}x-b_{1}\) and then

\[f_{k}(x)=W_{k}\sigma(f_{k-1}(x))-b_{k}\]

One can always choose \(W_{1},\ldots W_{l-2}\) and \(b_{1},\ldots b_{l-2}\) such that the components of \(f_{k}(x_{i})\) belongs to \(I\). Therefore, if we start with \(d\) distinct points \(x_{1},\ldots x_{d}\), we end up with \(d\) distinct outputs \(\sigma(f_{l-2}(x_{1})),\ldots\sigma(f_{l-2}(x_{d}))\) by the injectivity of \(\sigma\) on \(I\). Now for \(W_{l-1},W_{l},b_{l-1}\), and \(b_{l}\), we use the same construction as in Proposition 2.3.
With these settings described above, we have the following consequence.

**Theorem 2.5**.: _Let \((x_{i},y_{i})_{i=\overline{1,d}}\) be a data set with \(x_{i}\in\mathbb{R}^{p},y_{i}\in\mathbb{R}\), and the \(x_{i}\) are distinct. Assume that the activation function \(\sigma\) is smooth and satisfies Assumption 2.2. Let \(L\) be the mean squared loss function of a feedforward neural network with \(l\) hidden layers and with the last one of width \(h\geq d\). Then, the set \(M=L^{-1}(0)\) is generically (that is, possibly after an arbitrarily small change to the data set) a smooth \(n-d\) dimensional submanifold nonempty of \(\mathbb{R}^{n}\)._

Proof.: This is a consequence of Theorem 2.1 and Corollary 2.4. Notice that because \(\sigma^{\prime}\) is not constant, we can find an interval \(I\) such that \(\sigma^{\prime}\) has constant sign on that interval, thus is one-to-one as the Corollary 2.4 requires. 

## 3 Interpolation with polynomial activation function

If the activation function \(\sigma\) is a non-constant polynomial, the interpolation problem depends very much on the \(x_{i}\) and the degree of \(\sigma\). More precisely, we have the following result

**Proposition 3.1**.: _Let \((x_{i},y_{i})_{i=\overline{1,d}}\) be a data set with \(x_{i}\in\mathbb{R}^{p},y_{i}\in\mathbb{R}\), and the \(x_{i}\) are distinct. If \(\sigma\) is a polynomial of degree \(m\), then we have the following two statements_

1. _If_ \(d>\sum_{k=1}^{m}\binom{p+k}{k}\)_, then the interpolation problem in Proposition_ 2.3 _is not possible._
2. _If_ \(d\leq\sum_{k=1}^{m}\binom{p+k}{k}\) _and_ \((1,x_{i},x_{i}^{\otimes 2},\ldots,x_{i}^{\otimes m})_{i=\overline{1,d}}\) _are linearly independent, then the interpolation problem in Proposition_ 2.3 _is possible._

Proof.: Since the interpolation problem is equivalent to proving that the functions \(\sigma(w^{T}x_{i}-b)\) are linearly independent (over \(w\) and \(b\)), we will show that one can find nontrivial coefficients \((c_{i})_{i=\overline{1,d}}\) such that

\[\sum_{i=1}^{d}c_{i}\sigma(w^{T}x_{i}-b)=0, \tag{3}\]

for any \(w\in\mathbb{R}^{p}\) and \(b\in\mathbb{R}\). For simplicity, we can take \(b=0\) by considering \(x_{i}\leftarrow(x_{i},-1)\), and \(w\leftarrow(w,b)\). By these notations, equation (3) is equivalent to

\[\sum_{i=1}^{d}c_{i}\sigma(w^{T}x_{i})=0, \tag{4}\]

for any \(w\in\mathbb{R}^{p+1}\). Since \(\sigma\) is a polynomial of degree \(m\), equation (4) is equivalent to

\[\sum_{i=1}^{d}c_{i}(w^{T}x_{i})^{k}=0, \tag{5}\]

for any \(k=\overline{0,m}\) and \(w\in\mathbb{R}^{p+1}\). And equation (5) is equivalent to

\[\sum_{i=1}^{d}c_{i}x_{i}^{\otimes k}=0, \tag{6}\]

for any \(k=\overline{0,m}\). Thus, our problem is reduced to finding a linear combination of the elements \((1,x_{i},x_{i}^{\otimes 2},\ldots,x_{i}^{\otimes m})\).

It is well known that \(\text{Sim}^{k}(\mathbb{R}^{p+1})\), i.e. the space of symmetric tensors of order \(k\), is spanned by elements of the form \(v^{\otimes k}\) and has dimension \(\binom{p+k}{k}\). Consequently, if the number of data points \(x_{i}\) is bigger than the dimension of \(\bigoplus_{k=1}^{m}\text{Sim}^{k}(\mathbb{R}^{p+1})\), which is \(\sum_{k=1}^{m}\binom{p+k}{k}\), then we can find a linear dependence.

On the other hand, if \(d\leq\sum_{k=1}^{m}\binom{p+k}{k}\), and the vectors \((1,x_{i},x_{i}^{\otimes 2},\ldots,x_{i}^{\otimes m})_{i=\overline{1,d}}\) are linearly independent, then \(c_{i}\) must all be equal to \(0\), thus the interpolation is possible. 

As the following results in this section heavily rely on property 2 stated in Proposition 3.1, we introduce the following assumption.

**Assumption 3.2**.: _Let \((x_{i},y_{i})_{i=\overline{1,d}}\) be a data set with \(x_{i}\in\mathbb{R}^{p},y_{i}\in\mathbb{R}\). We require that \(d\leq\sum_{k=1}^{m}\binom{p+k}{k}\) and \((1,x_{i},x_{i}^{\otimes 2},\ldots,x_{i}^{\otimes m})_{i=\overline{1,d}}\) are linearly independent._

We have the following consequence from Proposition 3.1

**Corollary 3.3**.: _Let \((x_{i},y_{i})_{i=\overline{1,d}}\) be a data set and \(\sigma\) a polynomial function. Assume that the activation function \(\sigma\) has degree \(m\) and our data set satisfy Assumption 3.2. Then, in the family of feedforward neural nets of \(l\) hidden layers, with the last hidden layer having \(h\geq d\) neurons, and remaining hidden layers of any width, we can find one that interpolates our data set, i.e. \(f_{w,b}(x_{i})=y_{i}\) for all \(i\)._
Proof.: The proof follows a similar approach to that of Corollary 2.4. Notice that because \(\sigma\) is a non-constant polynomial function, we can find an interval \(I\) such that \(\sigma^{\prime}\) has a constant sign on that interval, thus is one-to-one as the Corollary 2.4 requires. 

With these settings described above, we have the following consequence

**Theorem 3.4**.: _Let \((x_{i},y_{i})_{i=\overline{1,d}}\) be a data set and \(\sigma\) a polynomial function. Assume that the activation function \(\sigma\) and our data set satisfy Assumption 3.2. Let \(L\) be the mean squared loss function of a feedforward neural network with \(l\) hidden layers and with the last one of width \(h\geq d\). Then, the set \(M=L^{-1}(0)\) is generically (that is, possibly after an arbitrarily small change to the data set) a smooth \(n-d\) dimensional submanifold nonempty of \(\mathbb{R}^{n}\)._

Proof.: This is a consequence of Proposition 3.1 and Corollary 3.3. 

## 4 The Hessian for the global minima

In this section, we describe the Hessian eigenspectrum of the loss function \(L\) evaluated at a point \(m\in M=L^{-1}(0)\). The following proposition is a result of [3], and it is true for any neural network architecture.

**Proposition 4.1**.: _Let \(M=L^{-1}(0)=\bigcap M_{i}\), where \(M_{i}=f_{i}^{-1}(0)\), be the locus of global minima of \(L\). If each \(M_{i}\) is a smooth codimension 1 submanifold of \(\mathbb{R}^{n}\), \(M\) is nonempty, and the \(M_{i}\) intersect transversally at every point of \(M\), then at every point \(m\in M\), the Hessian evaluated at \(m\) has \(d\) positive eigenvalues and \(n-d\) eigenvalues equal to 0._

Consider now a shallow neural net as in Theorem 2.5 or Theorem 3.4. Then we have the following Corrolary of Proposition 3.1 :

**Corollary 4.2**.: _Let \(L\) be the mean square loss function of a neural net as described above. Then, \(M\) is nonempty, and the Hessian of \(L\), evaluated at any point \(m\in M=L^{-1}(0)\) has \(d\) positive eigenvalues and \(n-d\) eigenvalues equal to 0._

Proof.: Without losing the generality, suppose our shallow neural network is in the setting of Theorem 2.5.

The locus of global minima \(M=L^{-1}(0)\) is the intersection of \(M_{i}\), where

\[M_{i}=\{(w,b)\in\mathbb{R}^{n}|f_{w,b}(x_{i})=y_{i}\}\]

Due to Proposition 3.1, it suffices to prove that \(M\) is non-empty, \(M_{i}\) are smooth of codimension 1, and that \(M_{i}\) intersects transversally at each point of \(M\).

The nonemptiness of \(M\) follows from Theorem 2.5. Each \(M_{i}\) is smooth of codimension 1, again by Theorem 2.5. for \(d=1\). It remains to prove that the intersection of \(M_{i}\) is transversal. Let \(m=(w,b)\in M\). We assume that the intersection at \(m\) is not transversal. This means the tangent space \(T_{m}M_{1}=T_{m}M_{i}\) for all \(i\). From our notations, we have that

\[f_{i}(w,b)=W_{2}\sigma(W_{1}x_{i}-b_{1})-b_{2}-y_{i},\]

The equality of the tangent spaces at \(m\), means that their normal vectors are collinear, i.e. \(\nabla f_{i}(w,b)=\alpha_{i}\nabla f_{1}(w,b)\) for some \(\alpha_{i}\in\mathbb{R}\). If we compute the partial derivatives with respect to \(W_{1},b_{1}\), and \(b_{2}\), we get

\[\frac{\partial f_{i}}{\partial W_{1}}(w,b)= -\frac{\partial f_{i}}{\partial b_{1}}(w,b)\otimes x_{i}\] \[\frac{\partial f_{i}}{\partial b_{2}}(w,b)= -1\]

From the partial derivative with respect to \(b_{2}\), we get that \(\alpha_{i}=1\) for all \(i\). Thus,

\[\frac{\partial f_{i}}{\partial b_{1}}(w,b)= \frac{\partial f_{j}}{\partial b_{1}}(w,b)\] \[\frac{\partial f_{i}}{\partial b_{1}}(w,b)\otimes x_{i}= \frac{\partial f_{j}}{\partial b_{1}}(w,b)\otimes x_{j}\]

for all \(i,j\). Since \(\sigma\) is smooth, we can find an interval \(I\) such that \(\sigma^{\prime}\) does not vanish on it. We consider a point \((w^{*},b^{*})\in\mathbb{R}^{n}\) such that all entries of \(W_{1}\) are equal to 0, all entries of \(W_{2}\) are different from 0, and all entries of \(-b_{1}\) belong to \(I\). With this setting, each component of \(\frac{\partial f_{i}}{\partial b_{1}}(w^{*},b^{*})\) is different from zero. So from the last two relations, we get \(x_{i}=x_{j}\) for all \(i,j\), which is a contradiction with the assumption of our data set.
Convergence to the global minima

In Section 2, we established the existence of an interpolation point. In this section, we present a method which probabilistically determines this point. This approach involves initializing the input-to-hidden weights randomly and optimizing the out-layer weights \(v\in\mathbb{R}^{h}\). This idea is inspired from [5]. Before we jump into the details, we will absorb the biases into the weights, simply adding to the inputs vectors \(x_{i}\) the \(p+1\) coordinate equal to \(1\). Thus in the rest of this section, we will assume that \(x_{i}\) is constructed this way and we will call this again \(x_{i}\) to keep the notations simple. Notice that the dimension of the vector changes now from \(p\) to \(p+1\).

Now, we need to minimize the loss function:

\[L(v):=\sum_{i=1}^{d}(v^{T}\sigma(Wx_{i})-y_{i})^{2}=||\sigma(XW^{T})v-y||^{2},\]

which is a simple linear regression problem. Moreover, if \(\sigma(XW^{T})\) has full rank, then the global minimum of this optimization problem is given by

\[\tilde{v}:=\phi^{T}(\phi\phi^{T})^{-1}y\]

where \(\phi:=\sigma(XW^{T})\). So the question we ask is how much overparameterization is needed to achieve a full rank for the matrix \(\sigma(XW^{T})\). Observe that

\[\phi\phi^{T}=\sigma(XW^{T})\sigma(XW^{T})^{T}=\sum_{l=1}^{h}\sigma(Xw_{l}) \sigma(Xw_{l})^{T}.\]

where \(w_{l}\) is the \(l\)-th line of \(W\). This leads us to the following definition.

**Definition 5.1**.: Let \(w\) be a random vector with a \(\mathcal{N}(0,I_{p+1})\) distribution. We define the following matrix

\[\tilde{\Sigma}(X):=\mathbb{E}_{w}[\sigma(Xw)\sigma(Xw)^{T}]\]

And let \(\tilde{\lambda}(X)\) be the minimum eigenvalue of \(\tilde{\Sigma}(X)\), i.e. \(\tilde{\lambda}(X):=\lambda_{min}(\tilde{\Sigma}(X))\)

The following Proposition is a consequence of the interpolation property.

**Proposition 5.2**.: _If the activation function \(\sigma\) and our data set \(\left(x_{i},y_{i}\right)_{i=\overline{1,d}}\) satisfies Assumption 2.2 or 3.2, then \(\tilde{\lambda}(X)>0\)._

Proof.: Let \(v\in\mathbb{R}^{d}\) such that \(v\tilde{\Sigma}(X)v^{T}=0\). This is equivalent to

\[\sum_{i=1}^{d}v_{i}\sigma(w^{T}x_{i})=0, \tag{7}\]

for almost every \(w\in\mathbb{R}^{p+1}\). If \(\sigma\) satisfies Assumption 2.2, then, using the same arguments as in Proposition 2.3, we get that \(v=0\). Otherwise, we use the reasoning from 3.1. Therefore, \(\tilde{\Sigma}(X)\) is a symmetric positive definite matrix. 

In [5], using matrix concentration inequalities and Gaussian Lipschitz concentration inequalities, one can prove the non-singularity of \(\phi\phi^{T}\) when the activation function \(\sigma\) has a bounded derivative. Using similar arguments as in [5], we extend this result for continuous activation functions \(\sigma\) which are not polynomials of degree less than \(d-2\).

We state here one result which plays the leading role in our arguments.

**Theorem 5.3**.: _(Matrix Chernoff) Let \(\left(A_{l}\right)_{l=\overline{1,l}}\) be sequence of independent, random, Hermitian matrices of dimension \(n\). Assume that \(0\preceq A_{l}\preceq R\cdot I_{n}\) for \(l=\overline{1,k}\). Then_

\[\mathbb{P}\left(\lambda_{min}\left(\sum_{l=1}^{k}A_{l}\right)\leq(1-\delta) \lambda_{min}\left(\sum_{l=1}^{k}\mathbb{E}(A_{l})\right)\right)\leq n\left( \frac{e^{-\delta}}{(1-\delta)^{1-\delta}}\right)^{\frac{\lambda_{min}\left( \sum_{l=1}^{k}\mathbb{E}(A_{l})\right)}{M}}\]

_for any \(\delta\in[0.1)\)_

Now we are ready for the main result of this section.
**Theorem 5.4**.: _Let \((x_{i},y_{i})_{i=\overline{1,d}}\) be a data set with \(x_{i}\in\mathbb{R}^{p+1},y_{i}\in\mathbb{R}\), and assume that \(x_{i}\) are distinct. Consider a shallow neural network with \(h\) hidden nodes of the form \(f(v,W):=v^{T}\sigma(Wx)\) with \(W\in\mathcal{M}_{h\times(p+1)}(\mathbb{R})\) and \(v\in\mathbb{R}^{h}\). Let \(\mu\) be the Gaussian measure. We assume the activation function \(\sigma\in C(\mathbb{R})\cap L^{2}(\mathbb{R},\mu)\) and is not a polynomial of degree less than \(d-2\). We initialize the entries of \(W\) with i.i.d. \(\mathcal{N}(0,1)\). Also, assume_

\[h\geq\frac{C_{\sigma}d\log(d)}{\tilde{\lambda}(X)}\]

_where \(C_{\sigma}\) is a constant that depends only on \(\sigma\). Then, the matrix \(\sigma(XW^{T})\) has full row rank with probability at least \(1-\frac{1}{d^{10\cdot\sigma}}\)._

Proof.: It suffices to prove that \(\sigma(XW^{T})\sigma(XW^{T})^{T}\) is non-singular with high probability. First, observe that

\[\phi\phi^{T}=\sigma(XW^{T})\sigma(XW^{T})^{T}=\sum_{l=1}^{h}\sigma(Xw_{l}) \sigma(Xw_{l})^{T}\geq\sum_{l=1}^{h}\sigma(Xw_{l})\sigma(Xw_{l})^{T}\mathbb{1 }_{\{[|\sigma(Xw_{l})|]<T_{d}\}}.\]

Here \(T_{d}\) is a function of \(d\) which will be determined later in the proof. Applying the Matrix Chernoff concentration inequality for \(A_{l}=\sigma(Xw_{l})\sigma(Xw_{l})^{T}\mathbb{1}_{\{[|\sigma(Xw_{l})|]<T_{d}\}}, R=T_{d}^{2}\) and \(\tilde{A}(w)=\sigma(Xw)\sigma(Xw)^{T}\mathbb{1}_{\{[|\sigma(Xw)|]<T_{d}\}}\), we get

\[\lambda_{min}\left(\phi\phi^{T}\right)\geq h(1-\delta)\lambda_{min}\left( \mathbb{E}\left[\tilde{A}(w)\right]\right) \tag{8}\]

holds with probability \(1-d\left(\frac{e^{-\delta}}{(1-\delta)^{1-\delta}}\right)^{\frac{h\lambda_{min }\left(\mathbb{E}[\tilde{A}(w)\right]\right)}{T_{d}^{2}}}\). We can fix \(\delta\) from now on, for instance we can pick \(\delta=1/2\).

Now, it remains to prove that \(\mathbb{E}[\tilde{A}(w)]\) is a positive definite matrix. Let \(v\in\mathbb{R}^{d}\) such that \(v\mathbb{E}[\tilde{A}(w)]v^{T}=0\). This is equivalent to

\[\sum_{i=1}^{d}v_{i}\sigma(w^{T}x_{i})=0, \tag{9}\]

for any \(w\in\mathbb{R}^{p+1}\) that satisfies almost surely \(\|\sigma(Xw)\|<T_{d}\). Because \(\sigma\) is continuous we actually have relation (9) valid for all \(w\) with \(\|\sigma(Xw)\|<T_{d}\).

We impose now a first condition on \(T_{d}\), namely, we require that \(\sigma(0)\sqrt{d}<T_{d}\). Since the \(x_{i}\) are distinct, we can find a vector \(w\in\mathbb{R}^{p+1}\) such that \(w^{T}x_{i}=t_{i}\) are distinct for all \(i\). We can take this \(w\) such that the last component is also \(0\). With the choice of \(T_{d}\), we can now scale \(w\) to be sufficiently small so that \(\|\sigma(Xw)\|_{2}<T_{d}\). Then for any \(a,b\in\mathbb{R}\) that satisfy \(\sum_{i=1}^{d}\sigma^{2}(at_{i}-b)<T_{d}^{2}\) we have

\[\sum_{i=1}^{d}v_{i}\sigma(at_{i}-b)=0. \tag{10}\]

Let \(\zeta\in C_{0}^{\infty}(\mathbb{R},[0,\infty))\), i.e. \(\zeta\) is non-negative, infinitely differentiable with compact support on \([-1,1]\) and \(\int_{\mathbb{R}}\zeta(x)dx=1\). We define for \(\epsilon>0\), the following function

\[\sigma_{\epsilon}(t)=\int_{\mathbb{R}}\frac{1}{\epsilon}\zeta\left(\frac{t-x}{ \epsilon}\right)\sigma(x)dx\]

Since \(\sigma\) is continuous, standard arguments show that

\[\sigma_{\epsilon}\xrightarrow[\epsilon\to 0]{u}\sigma \tag{11}\]

Since \(\sigma\) is not a polynomial of degree less than \(d-2\), we can find \(M>0\) large enough such that \(\sigma|_{[-M,M]}\) is not a polynomial of degree less than \(d-2\). From 11, we can find a small \(\epsilon\) such that \(\sigma_{\epsilon}\) is not a polynomial of degree less than \(d-2\) restricted to \([-M,M]\).

One of the key steps is the following

_Claim._ Since \(\sigma_{\epsilon}\) is not a polynomial of degree less than \(d-2\) on \([-M,M]\), we can find

\[b_{0}\in[-M,M]\text{ such that }\sigma_{\epsilon}^{(k)}(-b_{0})\neq 0\text{ for any }k=\overline{0,d-1}.\]

One way to justify this claim follows for instance from the argument indicated by Pinkus in [6] which actually refers to [4, Theorem of Agmon, page 53] with an easy adaptation.

Another way to see this is the following. Consider

\[D_{k}=\{b\in(-M,M):\sigma_{\epsilon}^{(k)}(-b)\neq 0\}\text{ for }k=0,1,\ldots,d-1.\]
In the first place, we notice \(D_{k}\) are open sets in \((-M,M)\) and \(D_{d-1}\neq\O\). By induction we can assume that \(D_{k+1}\cap D_{k+2}\cdots\cap D_{d-1}\neq\O\) and then taking a \(b\) in this intersection and \(\delta\) sufficiently small such that \(B(b,\delta)\subset D_{k+1}\cap D_{k+2}\cdots\cap D_{d-1}\). Now, we can argue that there must be \(b^{\prime}\in B(b,\delta)\) such that \(\sigma_{\epsilon}^{(k)}(-b^{\prime})\neq 0\) (otherwise \(\sigma_{\epsilon}^{(k+1)}(b)=0\)). Therefore, \(b^{\prime}\in D_{k}\cap D_{k+1}\cap D_{k+2}\cdots\cap D_{d-1}\) which shows by induction that \(D_{0}\cap D_{1}\cap\ldots,\cap D_{d-1}\neq\O\).

With the above Claim at hand, without loss of generality, we can consider \(M\) such that \((-b_{0}-\epsilon,-b_{0}+\epsilon)\subset[-M,M]\). Let \(T_{d}:=\sqrt{d}(\sup_{x\in[-M,M]}|\sigma(x)|+1)\). Notice that this \(T_{d}\) already satisfies the condition we require for the choice of \(w\) above. We can find a small interval \([-\rho,\rho]\) such that \(at_{i}-b_{0}-\epsilon z\in[-M,M]\) for any \(a\in[-\rho,\rho]\) and \(z\in[-1,1]\), and from the definition of \(T_{d}\) we have

\[\sum_{i=1}^{d}\sigma^{2}(at_{i}-b_{0}-\epsilon z)<T_{d}^{2} \tag{12}\]

for any \(a\in[-\rho,\rho]\) and \(z\in[-1,1]\). Consequently,

\[\sum_{i=1}^{d}v_{i}\sigma_{\epsilon}(at_{i}-b_{0})=\int_{\R}\zeta(z)\sum_{i=1} ^{d}v_{i}\sigma(at_{i}-b_{0}-\epsilon z)dz=0 \tag{13}\]

for any \(a\in[-\rho,\rho]\).

If we differentiate \(k\) times relation (13) with respect to \(a\), we get

\[\sum_{i=1}^{d}v_{i}t_{i}^{k}\sigma_{\epsilon}^{(k)}(at_{i}-b_{0})=0.\]

Taking \(a=0\) for each equation, we get a system of \(d\) equations

\[\sum_{i=1}^{d}v_{i}t_{i}^{k}=0, \tag{14}\]

for each \(k=\overline{0,d-1}\). Since the matrix system of (14) is a Vandermonde matrix, and the \(t_{i}\) are distinct, we get that all \(v_{i}\) must be equal to \(0\). Hence \(\mathbb{E}[\hat{A}(w)]\) is a symmetric positive definite matrix and \(\phi\) has full rank with probability at least \(1-de^{-\frac{\gamma\hat{\lambda}_{min}(\mathbb{E}[\hat{A}(w)])}{T_{d}^{2}}}\) where \(\gamma\) is a constant depending explicitly on \(\delta\). This probability is larger than \(1-\frac{1}{d^{200}}\) as long as

\[h\geq\frac{C_{\sigma}d\log(d)}{\lambda_{min}(\mathbb{E}[\hat{A}(w)])}\geq \frac{C_{\sigma}d\log(d)}{\hat{\lambda}(X)}\]

Following the same line of reasoning, we have a similar result for polynomial functions. More precisely, we have the following result

**Theorem 5.5**.: _Let \((x_{i},y_{i})_{i=\overline{1,d}}\) be a data set and \(\sigma\) a polynomial function. Assume that the activation function \(\sigma\) and our data set satisfy Assumption 3.2.Consider a shallow neural network with \(h\) hidden nodes of the form \(f(v,W):=v^{T}\sigma(Wx)\) with \(W\in\mathcal{M}_{h\times(p+1)}(\R)\) and \(v\in\R^{h}\). We initialize the entries of \(W\) with i.i.d. \(\mathcal{N}(0,1)\). Also, assume_

\[h\geq\frac{C_{\sigma}d\log(d)}{\hat{\lambda}(X)}\]

_where \(C_{\sigma}\) is a constant that depends only on \(\sigma\). Then, the matrix \(\sigma(XW^{T})\) has full row rank with probability at least \(1-\frac{1}{d^{100}}\)._

Proof.: Following the same reasoning as in Theorem 5.4, we get to the equation

\[\sum_{i=1}^{d}v_{i}\sigma(w^{T}x_{i})=0, \tag{15}\]

for any \(w\in\R^{p+1}\) that satisfies \(||\sigma(Xw)||<T_{d}\). Let \(T_{d}:=\sigma(0)\sqrt{d}+1\). Using the same arguments as in Proposition 3.1, equation 15 is equivalent to

\[\sum_{i=1}^{d}v_{i}x_{i}^{\otimes k}=0, \tag{16}\]

for any \(k=\overline{0,m}\). From assumption 3.2 we get that all \(v_{i}\) must be equal to \(0\). The rest follows as in Theorem 5.4.