Any linear operator $T$ satisfies $\lvert \lvert T x \rvert \rvert = \lvert \lvert T \rvert \rvert \cdot \lvert \lvert x \rvert \rvert$

Question

Let $(X,\lvert \lvert \cdot \rvert \rvert_{X})$ and $(Y,\lvert \lvert \cdot \rvert \rvert _{Y})$ and $X$ be finite dimensional, then show that any linear operator $T$ satisfies $\lvert \lvert T x \rvert \rvert = \lvert \lvert T \rvert \rvert \cdot \lvert \lvert x \rvert \rvert$ for some $x \neq 0$.

My idea:

I have shown that $T$ is bounded, now let us assume that for any $x \neq 0$ we have $\lvert \lvert Tx\rvert \rvert < \lvert \lvert T\rvert \rvert \cdot \lvert \lvert x \rvert \rvert$, then:

we have $\lvert \lvert T(\frac{x}{\lvert \lvert x \rvert \rvert})\rvert \rvert < \lvert \lvert T \rvert \rvert$ for all $x \neq 0$. But at the same time $\{ \frac{x}{\lvert \lvert x \rvert \rvert}: x \neq 0\}=\partial B_{1}(0)$ and $\lvert \lvert T \rvert \rvert=\sup\limits_{ x \in \partial B_{1}(0)}\lvert \lvert Tx\rvert \rvert_{Y}$

So we can construct a sequence $(x_{n})_{n}\subseteq \partial B_{1}(0)$ such that $\lim\limits_{n \to \infty}\lvert \lvert Tx_{n}\rvert \rvert _{Y}=\lvert \lvert T\rvert \rvert$

Since $\dim X<\infty$, we know that the closed unit ball $\overline{B_{1}^{X}(0)}$ is compact and hence $\partial B_{1}(0)$ is compact since it is closed and a subset. Thus there exists a convergent subsequence $(x_{n(k)})_{k \in \mathbb N}$ of $(x_{n})_{n \in \mathbb N}$ and $x \in \partial B_{1}(0)$ such that $\lim\limits_{k \to \infty}x_{n(k)}=x$. By the boundedness of $T$ and thus the continuity, it must follow that $\lim\limits_{k \to \infty}\lvert \lvert Tx_{n(k)}\rvert\rvert_{Y}=\lvert\lvert Tx\rvert\rvert_{Y}$ and thus

$$\lvert \lvert Tx\rvert \rvert _{Y}=\lim\limits_{k \to \infty}\lvert \lvert Tx_{n(k)}\rvert \rvert _{Y}=\lim\limits_{n \to \infty}\lvert \lvert Tx_{n}\rvert \rvert _{Y}=\lvert \lvert T\rvert \rvert$$

which is a contradiction since $x \in \partial B_{1}(0)$

Answer 1

Your idea is right, but the argument can be made simpler. Let $X_1=\{x:\ \|x\|=1\}$. Since $X$ is finite-dimensional, $X_1$ is compact. Consider the function $X_1\to[0,\infty)$ given by $x\longmapsto \|Tx\|$. Since $T$ is bounded and the norm is continuous, this is a continuous function. The image of the compact set $X_1$ is then compact; in particular there exists $x_0$ such that $\|Tx_0\|=\max\{\|Tx\|:\ \|x\|=1\}=\|T\|$.

Answer 2

I think your proof works fine.

One small thing, the initial assumption that $\|Tx\| < \|T\|\|x\|$ for all $x \neq 0$ seems to be unnecessary, so it would be cleaner to remove it, and simply take a sequence $x_n$ is the closed unit sphere that tends to the supremum.