Folland Real Analysis Exercise 5.15 - A unique linear tranformation $S$ such that $T = S \circ \pi$ where $\pi$ is the projection onto $X/N(T)$

Question

Consider the following problem, Exercise 15 of chapter 5 of Folland's Real Analysis, 2nd edition:

Suppose that $X$ and $Y$ are normed vector spaces and $T \in L(X, Y)$. Let $N(T) = \{x \in X \ : \ Tx = 0\}$.

a. Show that $N(T)$ is a closed subspace.

b. Show that there exists a unique $S \in L(X/N(T), Y)$ such that $T = S \circ \pi$ where $\pi: X \to X/N(T)$ is the canonical projection, and that $||S|| = ||T||$.

Part a. is clear. My question regards item b.. I tried defining $S(x + N(T)) = T(x)$. Is it really that trivial? I think I am missing something. Regarding the equality of norms, I was able to show that $||S|| \geq ||T||$: $$ \sup_{||x + N(T)|| \leq 1} |S(x + N(T)| = \sup_{||x + N(T)|| \leq 1} |T(x)| = \sup_{\inf_{y \in N(T)} ||x - y|| \leq 1} |T(x)| \geq||T||, $$ but I am stuck with the other inequality.

Any help will be the most appreciated.

Thanks in advance and kind regards.

Answer 1

I am stuck with the other inequality.

Given $t\in N(T)$ and $x\in X$, we have $$\|S(\pi(x))\|=\|T(x)\|=\|T(x+t)\|\leq \|T\|\|x+t\|$$ and thus $$\|S(\pi(x))\|\leq \inf_{t\in N(T)} \|T\|\|x+t\|=\|T\|\|\pi(x)\|,\quad\forall\ x\in X.$$ Therefore, $$\|S\|=\sup_{\substack{x\in X\\ x\neq 0}}\frac{S(\pi(x))}{\|\pi(x)\|}\leq \|T\|$$

Edit in response to the comment.

Existence. Define $S:X/N(T)\to X$ by $S(\pi(x))=T(x)$. Show that $S$ is well-defined. It is clear that $T=S\circ \pi$.

Uniqueness. Suppose that $\tilde{S}:X/N(T)\to X$ satisfies $T=\tilde{S}\circ \pi$. Then $$\tilde{S}(z)=\tilde{S}(\pi(x))=T(x)=S(\pi(x))=S(z),\quad\forall \ z=\pi(x)\in X/N(T)$$ and thus $\tilde{S}=S$.

Answer 2

Pick a sequence $y_n\in N(T)$ such that $\|x+y_n\|\to\|x+N(T)\|$. Then $$\|S(x+N(T))\|=\|Tx\|=\|T(x+y_n)\|\leq\|T\|\|x+y_n\|\to\|T\|\|x+N(T)\|$$ So $\|S\|\le\|T\|$.