of $F$ is $2^{m}-2^{m-k}$.

With similar arguments as in [@Pott Theorem 3], one can obtain the desired assertions. For the convenience of the readers, we give the proof here. If $|S|<2^{m-k}$, then there are at most $2^{m-k}-2$ nonzero elements $v$ for which $x\mapsto \langle v,F(x)\rangle$ is not bent. Due to Lemma \[key\], $S$ cannot meet all subspaces of dimension $m+1-(m-k)=k+1$, hence there exists some subspace $T$ of dimension $k+1$ disjoint from $S\backslash \{0\}$. Then we have $T\cap (S\backslash \{0\})=\emptyset$, and hence $T\subseteq \mathbb{F}_{2}^{m} \backslash  (S\backslash \{0\})$. Therefore $T\subseteq \{v \in\mathbb{F}_{2}^{m}: x\mapsto \langle v,F(x)\rangle \text{~is bent}\}\cup\{0\},$ that is, there is a vectorial bent function from $\mathbb{F}_{2}^{n}$ to $\mathbb{F}_{2}^{k+1}$, which is impossible due to Nyberg’s theorem [@Nyb] on vectorial bent function which says that vectorial bent $(n,m)$-functions exist if and only if $n$ is even and $m\leq \frac{n}{2}$.

It is well known that the algebraic degree of a bent function in $n$ variables is upper bounded by $n/2$. This upper bound is tight, and is valid for vectorial bent functions as well. In fact, it is also true for all $(n,m)$-functions with maximum number of bent components.

\[deg\] Let $n=2k$, $m\geq k$ be positive integers. Let $F$ be an $(n,m)$-function with maximum 