g^{2^{i}}})+(\beta g+\overline{\beta g}))}.\end{aligned}$$

If $\alpha+\overline{\alpha}=0$, say $\alpha \in \mathbb{F}_{2^{k}}$, let $\beta$ be such that $\beta+\overline{\beta}\neq 0$, then $W_{F^{i}}(\beta,\alpha)=0$ as $(\alpha+\overline{\alpha})h({\rm Tr}^{k}_{e}(g+\overline{g}))=(\beta+\overline{\beta})^{2^{i}}$ has no solution. It implies that $\langle \alpha ,F^{i}\rangle$ is not bent. Therefore, by Theorem \[number\], $F^{i}$ has the maximal number of bent components if and only if for any $\alpha\not\in \mathbb{F}_{2^{k}}$, $\langle \alpha ,F^{i}\rangle$ is bent.

Now suppose $\alpha\not\in \mathbb{F}_{2^{k}}$, then $(\alpha+\overline{\alpha})h({\rm Tr}^{k}_{e}(g+\overline{g}))=(\beta+\overline{\beta})^{2^{i}}$ is equivalent to $h({\rm Tr}^{k}_{e}(g+\overline{g}))=(\beta+\overline{\beta})^{2^{i}}(\alpha+\overline{\alpha})^{-1}$. If $e=k$ and $h$ is a permutation over $\mathbb{F}_{2^{e}}=\mathbb{F}_{2^{k}}$, then the equation above reduces to $g+\overline{g}=h^{-1}((\beta+\overline{\beta})^{2^{i}}(\alpha+\overline{\alpha})^{-1})$. However, the mapping $g\mapsto g+\overline{g}$ is a bijection from $\mathcal{G}$ to $\mathbb{F}_{2^{k}}$, hence there is exactly one solution $g$ to the equation. It indicates that $W_{F^{i}}(\beta,\alpha)=\pm 2^k$ for any $\beta$, and thus $\langle \alpha ,F^{i}\rangle$ is bent.

If $e<k$, then for any fixed $\alpha \not\in \mathbb{F}_{2^{k}}$, by the fact $\mathbb{F}_{2^{e}}\subsetneqq \mathbb{F}_{2^{k}}$ and $(\alpha+\overline{\alpha})^{-1}({\rm Tr}^{n}_{k}(x))^{2^{i}}$ is surjective from $\mathbb{F}_{2^{n}}$ to $\mathbb{F}_{2^{k}}$, there must exist $\beta_{0}\in \mathbb{F}_{2^{n}}$ such that $(\alpha+\overline{\alpha})^{-1}(\beta_{0}+\overline{\beta_{0}})^{2^{i}}\not\in \mathbb{F}_{2^{e}}$. However, $h$ is a map from $\mathbb{F}_{2^{e}}$ to $\mathbb{F}_{2^{e}}$, it implies that $W_{F^{i}}(\beta_{0},\alpha)=0$, and hence $\langle \alpha ,F^{i}\rangle$ is not bent. Therefore $F^{i}$ cannot have any bent component if $e<k$.

Finally suppose $e=k$, and $h$ is not a permutation over $\mathbb{F}_{2^{e}}$, the equation reduces to 