Linear function in normed vector space (Folland, Real Analysis Exercise)

Question

$\textbf{Exercise12}$ Let $X$ be a normed vector space and $M$ be a proper closed subspace of $X$.

(1) $\Vert x+M\Vert = \inf\{\Vert x+y \Vert : y\in M \} $ is a norm on $X/M$.

(2) For any $\epsilon>0$, there exists $x\in X $ such that $\Vert x \Vert =1 $ and $\Vert x+m \Vert \geq 1-\epsilon$.

(3) The projection map $\pi(x)=x+M$ from $X$ to $X/M$ has norm 1.

(4) If $X$ is complete, so is $X/M$.

I already proved Exercise 12. I stuck the following problem -(b).

$\textbf{Problem}$ Suppose that $X$ and $Y$ are normed vector spaces and $T \in L(X,Y)$ where $L(X,Y)$ be the space of all bounded linear maps from $X$ to $Y$. Let $N(T)=\{x\in X: Tx=0\}$. (Actually, $N(T)=\ker(T)$).

a. $N(T)$ is a closed subspace of $X$.

b. There is a unique $S \in L(X/N(T),Y) $ such that $T=S \circ \pi$ where $\pi :X \rightarrow X/N(T)$ is a projecion. Moreover, $\Vert S \Vert = \Vert T \Vert$ .

Firstly, I proved (a) in problem. Thus, I want to use the Exercise 12 for proving (b). However, I couldn't prove the uniqueness.

$\textbf{Attempt}$ Suppose $S_1,S_2 \in L(X/N(T),Y)$ such that $T=S_1\circ \pi$ and $T=S_2\circ \pi$. Then, $(S_1-S_2)\circ \pi = 0$. And, I have $\Vert \pi \Vert =1 $ ....

Any help is appreciated...

Thank you!

Answer 1

Your proof is almost complete. If $S_1\neq S_2$, then there is a $0\neq y \in X/N$ such that $S_1(y)\neq S_2(y)$. There is also $x\in X\setminus N$ such that $\pi (x)=y$. Then $S_1\circ \pi (x)\neq S_2\circ \pi (x)$ and thus the two operators $S_i\circ \pi$ are not equal (and thus not both equal to $T$), so $S$ is indeed unique.

Answer 2

The condition $T = S \circ \pi$ uniquely determines $S$:

$$S([x]) = S(\pi(x)) = Tx, \quad\forall x \in X$$

and this is indeed well-defined because $[x] = [y] \implies x-y \in N(T) \implies Tx = Ty$.

$S$ is clearly linear.

For the norm we have

$$\|T\| = \|S \circ \pi\| \le \|S\| \underbrace{\|\pi\|}_{= 1} = \|S\|$$

Conversely, for $x \in X$ we have $$\|S([x])\| = \|Tx\| = \|T(x+y)\| \le \|T\|\|x+y\|, \quad\forall y \in N(T)$$

so taking the infimum over $y \in N(T)$ gives

$$\|S([x])\| \le \|T\| \|[x]\|,\quad \forall x \in X \implies \|S\| \le \|T\|$$

We conclude $\|S\| = \|T\|$.