cover all the nontrivial functions in [@MZTZ; @Pott] according to the arguments of Remark \[cover\] below.

\[keythm-2\] Let $n=2k$ be a positive integer, $e$ a positive divisor of $k$. Let $F^{i}(x)=x^{2^{i}}h({\rm Tr}^{n}_{e}(x))$, where $i$ is a nonnegative integer, $h:\mathbb{F}_{2^{e}}\rightarrow \mathbb{F}_{2^{e}}$ is an arbitrary function. Then $F^{i}$ has the maximal number of bent components if and only if $e=k$ and $\pi$ is a permutation over $\mathbb{F}_{2^{e}}$. Moreover, when $e=k$ and $h$ is a permutation over $\mathbb{F}_{2^{e}}$, for any $\alpha \in \mathbb{F}_{2^{n}}$, ${\rm Tr}^{n}_{1}(\alpha F^{i})$ is a bent component of $F^{i}$ if and only if $\alpha \not\in \mathbb{F}_{2^{k}}$. For other cases, $F^i$ does not have any bent component.

For any $x\in\mathbb{F}_{2^{n}}$, write $x=g+y$ such that $y\in \mathbb{F}_{2^{k}}$ and $g\in\mathcal{G}$, where $\mathcal{G}$ is a set of representatives of cosets of $\mathbb{F}_{2^{k}}$ in $\mathbb{F}_{2^{n}}$, say $\mathbb{F}_{2^{n}}=\cup_{g\in \mathcal{G}}(g+\mathbb{F}_{2^{k}})$ and it is a disjoint union. Denote $x^{2^{k}}$ by $\overline{x}$, then we have for any $(\beta, \alpha) \in \mathbb{F}_{2^{n}}\times\mathbb{F}_{2^{n}}^*$ that $W_{F^{i}}(\beta,\alpha)$ equals $$\begin{aligned}
&&\sum\limits_{x\in \mathbb{F}_{2^{n}}}(-1)^{{\rm Tr}^{n}_{1}(\alpha F^{i}(x))+{\rm Tr}^{n}_{1}(\beta x)} \\
&=&\sum\limits_{x\in \mathbb{F}_{2^{n}}}(-1)^{{\rm Tr}^{k}_{1}({\rm Tr}^{n}_{k}(\alpha F^{i}(x)))+{\rm Tr}^{k}_{1}({\rm Tr}^{n}_{k}(\beta x))} \\
&=&\sum\limits_{g\in \mathcal{G}, y\in  \mathbb{F}_{2^{k}}}(-1)^{{\rm Tr}^{k}_{1}(h({\rm Tr}^{k}_{e}(g+\overline{g}))((\alpha g^{2^{i}}+\overline{\alpha g^{2^{i}}})+(\alpha + \overline{\alpha})y^{2^{i}}))+{\rm Tr}^{k}_{1}((\beta g+\overline{\beta g})+(\beta+\overline{\beta})y)}\\
&=&\sum\limits_{g\in\mathcal{G}}(-1)^{{\rm Tr}^{k}_{1}(h({\rm Tr}^{k}_{e}(g+\overline{g}))(\alpha g^{2^{i}}+\overline{\alpha g^{2^{i}}})+(\beta g+\overline{\beta g}))}\sum\limits_{y\in \mathbb{F}_{2^{k}}}(-1)^{{\rm Tr}^{k}_{1}(((\alpha+\overline{\alpha})h({\rm Tr}^{k}_{e}(g+\overline{g})))^{2^{k-i}}+(\beta+\overline{\beta}))y)}\\
&=&2^{k}\sum\limits_{(\alpha+\overline{\alpha})\pi({\rm Tr}^{k}_{e}(g+\overline{g}))=(\beta+\overline{\beta})^{2^{i}}}(-1)^{{\rm Tr}^{k}_{1}(h({\rm Tr}^{k}_{e}(g+\overline{g}))(\alpha g^{2^{i}}+\overline{\alpha 