any $(n,m)$-function is equal to $2^{m}-2^{m-k}$, and when the equality holds, the *algebraic degree* of such functions is at most $k$. For any vectorial bent $(n,k)$-function $F(x)$, with the inclusion map $i: \mathbb{F}^{k}_{2}\rightarrow \mathbb{F}^{m}_{2}$, $a\mapsto (a,0)$, it can also be regarded as an $(n,m)$-function which has the maximal number $2^{m}-2^{m-k}$ of bent components. For this reason, we call the $(n,m)$-function $G(x)=(F(x),0)$ *trivial*. Our main contributions to the constructions and characterizations of such functions are summarized as follows:

$\bullet$ The case of $(n,m)$-functions.

Based on the Maiorana-McFarland class, we present a nontrivial construction of such functions in [*bivariate representation*]{} $G(x,y)=(x^{2^{i}}\pi(y)+g(y),h(y))$, where $(x,y)\in \mathbb{F}_{2^{k}}\times \mathbb{F}_{2^{k}}$, $\pi: \mathbb{F}_{2^{k}}\rightarrow \mathbb{F}_{2^{k}}$ is a permutation, $g(y):\mathbb{F}_{2^{k}}\rightarrow \mathbb{F}_{2^{k}}$ and $h(y):\mathbb{F}_{2^{k}}\rightarrow \mathbb{F}_{2^{m-k}}$ are arbitrary functions.

$\bullet$ The case of $(n,n)$-functions.

1\) Let $F(x)=x^d$. We show that $F(x)$ has the maximal number of bent components if and only if $F(x)$ is a vectorial bent $(n,k)$-function. Say $F$ is trivial.

2\) For any positive integer $e$ dividing $k$, we consider the $(n,n)$-functions in [*univariate representation*]{} of the form $$F^{i}(x)=x^{2^{i}}h({\rm Tr}^{n}_{e}(x)),$$ where $0\leq i\leq n-1$, $h: \mathbb{F}_{2^{e}}\rightarrow \mathbb{F}_{2^{e}}$ is an arbitrary function. We show that the functions ${\rm Tr}^{n}_{1}(\alpha F^{i}(x))$ is bent for any $\alpha \in \mathbb{F}_{2^{n}}\setminus\mathbb{F}_{2^{k}}$ if and only if 