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In this section, we will determine the tight upper bound for the number of bent components of general $(n,m)$-functions, where $n$ is even and $m\geq \frac{n}{2}$, and present a trivial construction of $(n,m)$-functions which reach this bound.

[@Bose-Burton] \[Bose-Burton\] Let $\mathbb{F}_{q}$ be the finite field of order $q=p^s$ and $PG(n,q)$ the projective space of dimension $n$ over $\mathbb{F}_{q}$. A set $S$ of points in $PG(n,q)$ that meets all $(n-k)$-dimensional subspaces of $PG(n,q)$ has at least $\frac{q^{k+1}-1}{q-1}$ points with equality if and only if $S$ is a subspace of dimension $k$.

As it has been pointed out by the authors of [@Pott], one can get the following Lemma when applying Theorem \[Bose-Burton\] to the case of $p=2$.

\[key\][@Pott Corollary 1]. A set $S$ of elements in $\mathbb{F}_{2}^{n}\backslash\{0\}$ meeting all $(n+1-k)$-dimensional subspaces of $\mathbb{F}_{2}^{n}$ has at least $2^k-1$ elements with equality if and only if $S\cup \{0\}$ is a $k$-dimensional subspace of $\mathbb{F}_{2}^{n}$.

\[number\] Let $n=2k$, $m$ be positive integers with $m\geq k$, and $F$ an $(n,m)$-function. Denote by $S=\{v \in\mathbb{F}_{2}^{m}: x\mapsto \langle v,F(x)\rangle \text{is not bent}\}.$ Then $|S|\geq 2^{m-k}$, with equality if and only if $S$ is an $(m-k)$-dimensional subspace of $\mathbb{F}_{2}^{m}$. In particular, the maximal number of bent components 