Mulitplicativity of $\Vert \cdot\Vert_{L^\infty}$

Question

Let $(X,\mathscr{A},\mu)$ be some measure space and let $L^\infty(X)$ denote $L^\infty(X,\mathscr{A},\mu)$. Furthermore let $f\in L^\infty(X)\cap L^1(X,\mathscr{A},\mu)$ i.e. $\int_X \vert f\vert \ d \mu <\infty$ and $\operatorname{ess sup}_X \vert f \vert <\infty$. For some $k\in \mathbb{R}^+$ is the following true:
$\Vert f^k\Vert_{L^\infty(X)} \le \Vert f\Vert_{L^\infty(X)}^k$? Where $\Vert g\Vert_{L^\infty(X)}:= \operatorname{ess sup}_X \vert g\vert$ for any $g\in L^\infty(X)$.

Answer 1

Yes. We may assume $\|f\|_{\infty}=1$, and since $|x|^k\leq 1$ $\forall x\in [0,1]$ and every $k>0$, we get $\|f^k\|_{\infty}\leq 1$. You don't need to assume $f\in L_1$

Answer 2

I think I might have figured out all the details thanks to all of you. So for future reference, I'll share the answer. Let $f\in L^\infty(X)$, then we may assume without loss of generality that $\Vert f\Vert_{L^\infty}=1$. Thus, for every $\epsilon >0$ there exists some set $N\subset X$ with measure zero, such that $$ \sup_{x\in X\setminus N} \vert f(x) \vert -\epsilon \le 1 \implies \sup_{x\in X\setminus N}\vert f(x)\vert \le 1+\epsilon $$ Therefore we get that for any $k\in \mathbb{R}^+$ $$ (\sup_{x\in X\setminus N} \vert f(x)\vert )^k \le (1+\epsilon)^k $$ and since $\epsilon >0$ has been chosen arbirtrarily we have that $(\sup_{x\in X\setminus N} \vert f(x)\vert)^k \le 1$. Since we know that $(\sup_{x\in X\setminus N} \vert f(x)\vert)^k \ge \sup_{x\in X\setminus N}\vert f(x)\vert ^k$, we obtain that $\Vert f^k\Vert_{L^\infty}\le 1.$

Note: I do not claim the correctness of this proof