It is obvious that $$\sup_{[0,T]}{\left(|f|^2\right)}\leq (\sup_{[0,T]}|f|)^2.$$ My questions is when does the equality hold?
Progress: If $|f(x)|$ is not the zero function on $[0,T]$, the equality could hold. Let $m=\sup{|f|}$, $\forall \varepsilon >0$, $\exists x_0 \in[0,T]$, s.t. $$f(x_0)>m-\varepsilon>0 $$. So, $\sup{|f(x)|^2}>(m-\varepsilon )^2$. On the other hand, $\sup{|f(x)|^2}\leq m^2$. Hence, $$\sup{|f(x)|^2}=m^2$$ Am I right?
Equality will always hold.
Let $M = \sup_{[0,T]} |f|$. For any $\epsilon > 0$, there is some $t \in [0,T]$ with $|f(t)| > M - \epsilon$.
It follows that $|f(t)|^2 > (M - \epsilon)^2$. So, we have $\sup_{[0,T]} |f|^2 \geq (M-\epsilon)^2$
Thus, we have $$ \sup_{[0,T]} |f|^2 \geq \sup_{\epsilon > 0}(M-\epsilon)^2 = M^2 $$ That is, we have $$ \sup_{[0,T]} |f|^2 \geq [\sup_{[0,T]}|f|]^2 $$ which is the opposite direction of the "obvious" equality. The conclusion follows.