How to prove the inequality of the following expression [on hold]

Which one of these is have the large value?
Sup{||f(t)||2:t∈[a,b]}Sup \{||f(t)||^2:t\in [a, b]\} and (Sup{||f(t)||:t∈[a,b]})2?\big( Sup\{||f(t)||:t \in [a, b]\}\big)^2? What is the proof?

=================

=================

1 Answer
1

=================

Observe
‖f(t)‖2=(‖f(t)‖)2≤(supτ∈[a,b]‖f(τ)‖)2\begin{align}
\|f(t)\|^2 = \left(\|f(t)\| \right)^2 \leq \left(\sup_{\tau\in [a, b]} \|f(\tau)\|\right)^2
\end{align}
which holds for all t \in [a, b]t \in [a, b], then it follows
\begin{align}
\sup_{t\in [a, b]} \|f(t)\|^2 \leq \left(\sup_{\tau\in [a, b]} \|f(\tau)\| \right)^2.
\end{align}\begin{align}
\sup_{t\in [a, b]} \|f(t)\|^2 \leq \left(\sup_{\tau\in [a, b]} \|f(\tau)\| \right)^2.
\end{align}

Similarly, we have that
\begin{align}
(\|f(t)\|)^2 \leq \|f(t)\|^2 \leq \sup_{\tau \in [a, b]}\|f(\tau)\|^2
\end{align}\begin{align}
(\|f(t)\|)^2 \leq \|f(t)\|^2 \leq \sup_{\tau \in [a, b]}\|f(\tau)\|^2
\end{align}
which holds for all t \in [a, b]t \in [a, b] which means
\begin{align}
\left(\sup_{t \in [a, b]}\|f(t)\|\right)^2 \leq \sup_{\tau \in [a, b]}\|f(\tau)\|^2.
\end{align}\begin{align}
\left(\sup_{t \in [a, b]}\|f(t)\|\right)^2 \leq \sup_{\tau \in [a, b]}\|f(\tau)\|^2.
\end{align}
Hence we have that
\begin{align}
\left(\sup_{t \in [a, b]}\|f(t)\|\right)^2 = \sup_{\tau \in [a, b]}\|f(\tau)\|^2.
\end{align}\begin{align}
\left(\sup_{t \in [a, b]}\|f(t)\|\right)^2 = \sup_{\tau \in [a, b]}\|f(\tau)\|^2.
\end{align}