Verify $||T||_{L(X,Y)}$ norm is given by $||T||_{L(X,Y)} = \sup_{x \in X, ||x||_X = 1} ||T(x)||_Y$.

Question

Can somebody help me verify the problem

If $L(X,Y)$ is the set of all bounded and linear operator, then for every $T$ which is linear and bounded, $||T||_{L(X,Y)}$ norm is given by $||T||_{L(X,Y)} = \sup_{x \in X - \{0\}, ||x||_X \leq 1} \frac{||T(x)||_Y}{||x||_X} = \sup_{x \in X, ||x||_X = 1} ||T(x)||_Y$.

I don't know where to start with that.

Answer 1

If $x \in X, x \ne 0$, then $u := \frac1{\|x\|_X}x \in X$ and $\|u\|_X = \frac1{\|x\|_X}\|x\|_X = 1$. And $$\|T(u)\|_Y = \left\|T\left(\frac x{\|x\|_X}\right)\right\|_Y = \frac{\left\|T\left(x\right)\right\|_Y}{\|x\|_X}$$ since $T$ is linear and norms commute with positive multipliers.

So every value in $$S_1 := \left\{\frac{\|T(x)\|_Y}{\|x\|_X}\ \middle|\ x \in X, \|x\|_X \le 1, x\ne 0\right\}$$ is also in $$S_2 := \left\{\|T(u)\|_Y\mid u \in X, \|u\|_X = 1\right\}$$ I.e., $S_1 \subset S_2$. And manifestly $S_2 \subset S_1$, so in fact $S_1 = S_2$. And being the same set, they have the same supremum.

Note that this same argument also works on $$S_3 = \left\{\frac{\|T(x)\|_Y}{\|x\|_X}\ \middle|\ x \in X, x\ne 0\right\}$$ So the $\|x\|_X \le 1$ part of the first supremum was pointless.

As for why $\|T\|_{L(X, Y)}$ is the same value, that is because this is the definition of $\|T\|_{L(X, Y)}$.