$h(g+\overline{g})=(\beta+\overline{\beta})^{2^{i}}(\alpha+\overline{\alpha})^{-1}$. However, there must exist $\gamma \in \mathbb{F}_{2^{k}}$ such that its pre-image set $h^{-1}(\gamma)$ is empty. Then by the arguments as above, for any fixed $\alpha \not\in \mathbb{F}_{2^{k}}$, there must exist $\beta\in \mathbb{F}_{2^{n}}$ such that $(\alpha+\overline{\alpha})^{-1}(\beta+\overline{\beta})^{2^{i}}=\gamma$, and then $W_{F^{i}}(\beta,\alpha)=0$. Therefore $F^{i}$ cannot have any bent component.

Let $n=2k$, Proposition 3 in [@Pott] says that for an $(n,n)$-function $F$, if ${\rm Tr}^{n}_{1}(\alpha F(x))$ is a bent function for any $\alpha\in \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}}$, then ${\rm Tr}^{n}_{k}(\alpha F(x))$ is a vectorial bent function for any $\alpha\in \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}}$. By this observation together with Theorem \[keythm-2\], one can get the following result.

Let $n=2k$ be a positive integer, $i$ a nonnegative integer. Let $h$ be a permutation over $\mathbb{F}_{2^{k}}$. Then the function ${\rm Tr}^{n}_{k}(\alpha x^{2^{i}}h({\rm Tr}^{n}_{k}(x)))$ is vectorial bent for any $\alpha \not\in \mathbb{F}_{2^{k}}$.

The function $F^{i}(x)=x^{2^{i}}h({\rm Tr}^{n}_{k}(x))$, where $h$ is a permutation over $\mathbb{F}_{2^{k}}$, can possess a high algebraic degree. This class of functions provides the first nontrivial example of nonquadratic functions in univariate form with the maximal number of bent components. Some of such functions can reach the optimal algebraic degree $k$. For example, the algebraic degree of $x({\rm Tr}^{n}_{k}(x))^{2^{k-1}-1}$ is equal to $k$.

\[cover\] Let $h(x)=x$ for any $x\in \mathbb{F}_{2^{k}}$. Then 