Let $H$, $K$ be Banach spaces, and let $A: H \rightarrow K$ be a
bounded linear transformation.
Its norm is defined by:
\begin{equation}
\|A\| = sup\{\|Ah\|_K: \|h\|_H \le 1\}
\end{equation}
How to prove that $sup\{\|Ah\|_K: \|h\|_H \le 1\}$ is equivalent to:
\begin{equation}
sup\{\|Ah\|_K: \|h\|_H = 1\}
\end{equation}
Many thanks in advance,
--
Cesar
Let's assume $h^*$ maximizes (supremizes) $\|Ah\|_K$ and $\|h^*\|_H<1$, Then you can find scalar $\alpha$ such that:
$$\|h^*{'}\|_H=\|\alpha h^*\|_H=\alpha\|h^*\|_H=1$$
Since $\|h^*\|_H<1$, we should have $\alpha=\frac{1}{\|h^*\|_H}>1$ in order to hold the equality, therefore:
$$\|Ah^*{'}\|_K=\|A(\alpha h^*)\|_K=\alpha\|A h^*\|_K\geq \|Ah^*\|_K$$
Therefore, for any solution $h$ where $\|h\|<1$, there exist a solution $h{'}$ where $\|h{'}\|=1$, where the value of $\|Ah{'}\|_K$ is no less than the value of $\|Ah\|_K$.
As a result $h^*$ should have been equal to $h^*{'}$ in the first place, because the $\sup$ operator would return the larger value among the plausible ones.