$\alpha \in \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}}$, ${\rm Tr}^{n}_{1}(\alpha F(x))$ is bent [@Li2013 Theorem 2]. Thus, by Theorem \[number\], $F$ has $2^{n}-2^k$ bent components, and ${\rm Tr}^{n}_{1}(\alpha F(x))$ is bent if and only if $\alpha \in \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}}$. One will see that $F(x)$ is in fact a vectorial bent $(n,k)$-function. Indeed, for $1\leq i\leq 2^{r}-1$ denote $(i2^{k-r}+1)(2^k-1)+1$ by $d_{i}$, one has $F(x)=x^{(2^{k-1}+1)(2^k-1)+1}+\sum\limits^{2^{r-1}-1}_{i=1}(x^{d_{i}}+x^{d_{i}2^{k}})$. Note that $2(1+(2^{k-1}+1)(2^k-1))\equiv 2^k+1~{\rm mod}~(2^{2k}-1)$, the element $F(x)$ belongs to $\mathbb{F}_{2^k}$ for any $x\in \mathbb{F}_{2^n}$. Choosing $\alpha\in \mathbb{F}_{2^{n}}$ such that $\alpha+\alpha^{2^k}=1$, then ${\rm Tr}^{n}_{k}(\alpha F(x))=F(x)$ is a vectorial bent function by [@Pott Proposition 3]. This is why $F(x)$ can have the maximal number of bent components if one had known that $F(x)$ is a vectorial bent $(n,k)$-function in advance, and in our words $F(x)$ is trivial.

Below we will examine the case of the power $(n,n)$-functions, and give an observation on when it admits the maximal number of bent components. It turns out that if this is the case, it is trivial, i.e., it must be a vectorial bent $(n,k)$-function. Note that for any such $(n,n)$-function $F(x)$, let $S(F)=\{v\in \mathbb{F}_{2^{n}}~|~{\rm Tr}^{n}_{1}(v F(x)) \text{ is not bent} \}$, $S$ is a $k$-dimensional subspace of $\mathbb{F}_{2^{n}}$. Then by the fact that $S$ is 