Square root of the max is the max of the square root? I apologize if the question seems so obvious, but I don't have a strong base on maths nor I know the tools to prove this simple statement.
For a given function $f(x)$, is it true that
$$ \left(\max |f(x)|^2 \right)^{\frac{1}{2}} = \max |f(x)|.$$
I wonder if it should have a less or equal sign instead, but I do not know where to start.
Thanks in advance. 
Edit: I see that I was misunderstanding the answers... I thought my claim was false and should be $\leq$ instead. Now I see that equality holds by proving the inequalities in both ways. Thanks a lot!
 A: It's a big assumption that $\max |f(x)|$ or $\max(|f(x)|^2)$ exist but if one or the other does, they both do $(\max(|f(x)|^2))^{\frac 12} = \max |f(x)|$.
Suppose $\max |f(x)|$ exist.  That means there is $a\in \mathbb R$ so that for every $y\in \mathbb R$ we have $|f(y)| \le |f(a)|$ and $\max|f(x)| = |f(a)|$.
If $|f(y)|\le |f(a)|$ then $|f(y)|^2 \le |f(a)|^2$ so $\max(|f(x)|^2) = |f(a)|^2$.
And $(\max(|f(x)|^2)^{\frac 12} = (|f(a)|^2)^{\frac 12} = |f(a)| = \max (|f(x)|)$.
And the other direction is too similar to be worth dealing with.
Now a more subtle question is if $\max(|f(x)|)$ doesn't exist but $\sup(|f(x)|)$ does.
Does $\sup (|f(x)|^2)$ exist and if so does $(\sup(|f(x)|^2)^{\frac 12} = \sup |f(x)|$? 
The answer is still yes.
$|f(y)| \le \sup |f(x)| \iff |f(y)|^2 \le (\sup |f(x)|)^2$ so $\{|f(x)|^2\}$ is bounded above by $(\sup |f(x)|)^2|$.  So $\sup(|f(x)|^2)$ exists.
If $0< k < (\sup |f(x)|)^2$ then $\sqrt k < \sup|f(x)|$ and so there $y: \sqrt k < |f(y)| \le \sup |f(x)|$ and $k < |f(y)|^2$ so $k$ is not an upper bound.  So $\sup(|f(x)|^2) = (\sup(|f(x)|)^2$.
And so $(\sup(|f(x)|^2))^{\frac 12} =((\sup(|f(x)|)^2)^{\frac 12} = \sup(|f(x)|)$.
A: Short answer:
$$|a|>|b|\iff a^2>b^2.$$
A: Let $D$ be the domain of $f$. Pick a point $x \in D$. Then $|f(x)| \le \max_{y \in D} |f(y)|$ so that $|f(x)|^2 \le (\max_{y \in D} |f(y)|)^2$. Now take the maximum over all $x$ to find
$$\max_{x \in D} |f(x)|^2 \le (\max_{y \in D} |f(y)|)^2.$$
Again let $x \in X$. Then $|f(x)|^2 \le \max_{y \in D}|f(y)|^2$ so that $|f(x)| \le \left( \max_{y \in D} |f(y)|^2 \right)^{1/2}.$  Now take the maximum over all $x \in D$ to find
$$ \max_{x \in D} |f(x)| \le \left( \max_{y \in D}  |f(y)|^2 \right)^{1/2}.$$
A: Maximizing a function and maximizing any increasing function of that function are the same problem.
Let $g$ be any monotonically increasing function, so $x \leq y \implies g(x) \leq g(y)$.  Then if $x_0$ maximizes $f(x)$, $x_0$ must also maximize $g(f(x))$.  For if not, there is $y_0$ with $g(f(y_0)) > g(f(x_0))$. But that implies $f(y_0) > f(x_0)$, contradiction.
Thus, $\max(f(g(x)) = f(\max(g(x))$ (whenever $\max g(x)$ exists). Both are equal to $f(g(x_0))$ for any maximizer $x_0$ of $g(x)$.
In your case, both $g(x) = x^2$ and $g(x) = x^{1/2}$ are increasing (for $x>0$) so you can feel free to interchange these functions with the $\max$ operator.  Just pull the $\max$ outside and you are done.
