Supremum of norms of images of unit ball equals infimum of upper bounds of norms of images in the whole space

Question

For a linear operator $T \in B(X,Y)$, there exists

• $k \in \mathbb{R}_{>0}$ such that for all $x$ in the unit ball of $X: \lVert T(x)\rVert \leq k $;
• $k \in \mathbb{R}_{>0}$ such that for all $x \in X: \lVert T(x)\rVert \leq k \lVert x\rVert$.

This is supposed to imply that $$\text{sup}\{\lVert T(x) \rVert \mid \lVert x\rVert \leq 1 \} = \text{inf}\{k \in \mathbb{R}_{>0} \mid (\forall x \in X)(\lVert T(x)\rVert \leq k \lVert x\rVert)\}, $$ but this is not proved. I'd like to add rigour to the statement.

Also, the above should imply that for all $y \in X$ $$\lVert T(y)\rVert \leq \text{sup}\{\lVert T(x) \rVert \mid \lVert x\rVert \leq 1 \} \lVert y\rVert. $$

Answer 1

Take any $k\in \{k \in \mathbb{R}_{>0} \mid (\forall x \in X)(\lVert T(x)\rVert \leq k \lVert x\rVert)\}$. If $\|x\|\leq1$, then $$ \|Tx\|\leq k\|x\|=k. $$ As we can do this for all such $k$, we get $$ \|Tx\|\leq \inf\{k \in \mathbb{R}_{>0} \mid (\forall x \in X)(\lVert T(x)\rVert \leq k \lVert x\rVert)\}. $$ And we can do this for any $x$ with $\|x\|\leq1$, so $$ \sup\{\lVert T(x) \rVert \mid \lVert x\rVert \leq 1 \} \leq \inf\{k \in \mathbb{R}_{>0} \mid (\forall x \in X)(\lVert T(x)\rVert \leq k \lVert x\rVert)\}. $$ Also, for any nonzero $x\in X$, $$ \|Tx\|=\|T\left(\tfrac x{\|x\|}\right)\|\,\|x\|\leq \sup\{\lVert T(x) \rVert \mid \lVert x\rVert \leq 1 \}\,\|x\|. $$ Thus $$ \inf\{k \in \mathbb{R}_{>0} \mid (\forall x \in X)(\lVert T(x)\rVert \leq k \lVert x\rVert)\}\leq \sup\{\lVert T(x) \rVert \mid \lVert x\rVert \leq 1 \}. $$