# When is $f(|x|) \leq |f(x)|?$

When is $f(|x|) \leq |f(x)|?$
If $f(|x|) \leq |f(x)|$, what does this say about $f$?
Where $f$ is a real-valued function

I don't have much input on this, but I thought of this from having to show:
$$\sinh (|y|) \leq |\sin z|$$
for $z:= x+iy$ where I had
$$|\sin z| = \sqrt{\sin ^2x + \sinh ^2 y} \geq \sqrt{ \sinh ^2 y} = |\sinh y|$$
which suggests that $|\sinh y| \geq \sinh |y|$ for all $y\in \mathbb{R}$.

• Are there two questions (regarding necessary conditions and sufficient conditions separately) or only one question as the title says? If there is no mistake, the two questions marks denote two questions. – Megadeth Aug 16 '17 at 12:10

Nonnegative $x$:
For nonnegative values of $x$, you have $f(|x|)=f(x)$ which means that $f(|x|)\leq|f(x)|$ if and only if $f(x)\leq |f(x)|$ which is always true.
negative $x$:
For negative values of $x$, the inequality $f(|x|)\leq |f(x)|$ translates to $$f(-x) \leq |f(x)|$$ which, for example, is true for odd functions like $\sinh$ since if $f(-x)=-f(x)$, then $f(-x)=-f(x)\leq |f(x)|$.