I'm doing this exercise
Let $(E, |\cdot|)$ be a normed linear space and $M$ a closed subspace of $E$. Consider the quotient map $$T: E \to E/M, x \mapsto \hat x := x+M.$$ We endow $E/M$ with the quotient norm $\| \cdot \|$ defined by $$\| \hat x\| := d(x, M), \quad \forall x\in E.$$ Then $T$ is linear surjective. Prove that $\|T\| = 1$.
My attempt: We need the following Riesz's lemma.
Let $(E, |\cdot|)$ be a normed linear space and $M$ a closed proper subspace of $E$. Let $0 < \alpha < 1$. Then there exists an $x \in E$ with $|x|=1$ such that $|x-y| \ge \alpha$ for all $y \in M$. [A proof is given here]
First, we have $$\|Tx\| = \inf_{y\in M} |x-y| \le |x-0| = |x|.$$ It follows that $\|T\| \le 1$. By Riesz's lemma, for each $0<\varepsilon <1$, there is $x_\varepsilon \in E$ such that $|x_\varepsilon|=1$ and $\inf_{y\in M} |x_\varepsilon-y| \ge \varepsilon$. It follows that $$\|T x_\varepsilon\| \ge \varepsilon |x_\varepsilon|, \quad \forall \varepsilon \in (0, 1).$$ Take the limit $\varepsilon \to 1$, we get $\|T\| \ge 1$. This completes the proof.
Best Answer
Your proof is correct except for one minor thing (I'm sure you know this though):
The limit $\lim_{\epsilon \to 0} \|T x_\epsilon\|$ doesn't need to exist, so when you say "Take the limit $\epsilon \to 1$..." you actually should apply this to the inequality $\|T\|\ge \epsilon$ (this follows from your estimate).