Simple property of linear bounded maps

Question

Let $B(X,Y)$ be the space of bounded linear maps from normed space $X$ to normed space $Y$. If $T:X \to Y$, $T\in B(X,Y)$ we define a norm on $B(X,Y)$ by $$\|T\|=\sup\{\|T(x)\|_Y:\|X\|_X \le 1 \}$$ where $\|\cdot\|_X,\|\cdot\|_Y$ the norms on $X$ and $Y$ respectively.

Prove the following:
1) $\|T(x)\| \le \|T\|\cdot \|x\|$ for all $x\in X$
2) $\|T\|=\inf\{M\gt 0:\|T(x)\|\le M \cdot \|x\|$ for all $x\in X \}$

We already know from the fact that $T$ is bounded that there exists a positive $M$ such that $\|T(x)\|\le M \cdot \|x\|$ for all $x\in X$.

My issue with (1) is, that in the case that $\|x\|\le 1$, wouldn't multiplying $\|T\|$ with $\|x\|$ give a number potentially smaller than $T(x)$ for some of the $x$?

Answer 1

(1) Fix $x\in X$. If $x=0$, then the inequality is trivial. Otherwise, consider $z=x/\|x\|$. Then, $\|z\|=1$, so, $\|T(z)\|\leq\|T\|$. Hence, $\frac{1}{\|x\|}\|T(x)\|\leq\|T\|$, which is (1).

(2) Let $A=\{M>0:\|T(x)\|\leq M\|x\|\text{ for all }x\in X\}$. We want to prove that $\|T\|=\inf A$. Suppose $M\in A$. Fix $x\in X$ with $\|x\|\leq 1$. Then, $\|T(x)\|\leq M\|x\|\leq M$. Since this holds for all $x$ with $\|x\|\leq 1$, it follows that $\|T\|\leq M$. Furthermore, since tince holds for arbitrary $M\in A$, it follows that $\|T\|\leq\inf A$. If $\|T\|<\inf A$, then $\|T\|\notin A$, so there must be an $x\in X$ such that $\|T(x)\|> \|T\|\|x\|$, which contradicts (1). Therefore, $\|T\|=\inf A$.

Answer 2

Hint. $$\frac{‖Tx‖}{‖x‖} = \left\|T\bigg(\frac{x}{‖x‖}\bigg)\right\| \leq … $$

Answer 3

$(1)$ Let $x \in X \setminus \{0\}$ and define $\widetilde{x} :=\frac{x}{\|x\|_{X}}$ with $\|\widetilde{x}\|_{X}=1$. Then

$$\|T(x)\|_{Y} = \Big\|\tfrac{1}{\|x\|_{X}}T(x)\Big\|_{Y} \ \|x\|_{X} = \Big\|T\big(\tfrac{x}{\|x\|_{X}}\big)\Big\|_{Y} \ \|x\|_{X}= \|T(\widetilde{x})\|_Y \ \|x\|_{X} \leq \|T\| \ \|x\|_{X}.$$

$(2)$ Define $$ I := \inf\{M > 0 \colon \|T(x)\|_Y \leq M \ \|x\|_{X} \text{ for all } x \in X\}. $$ By $(1)$ we know that $$ \|T(x)\|_{Y} \leq \|T\| \ \|x\|_{X} $$ for all $x \in X$ and hence $I \leq \|T\|$. Now it remains to show that $\|T\| \leq I$. Let $x \in X$ with $\|x\|_{X}=1$. Then $$ \|T(x)\|_{Y} \leq I \ \|x\|_{X} = I. $$ Taking the supremum over all $x \in X$ with $\|x\|_{X}=1$ yields $\|T\| \leq I$.