Is the operator norm of the canonical surjection $X\to X/U$ equal to $1$? Let $X$ be a normed space, $U$ be a closed subspace of $X$ and $\pi:X\to X/U$ denote the canonical projection. By definition, $\pi$ is surjective and it is easy to show that $\pi\left(B^X_1(0)\right)=B^{X/U}_1(0)$, where $B^X_1(0)$ and $B^{X/U}_1(0)$ denote the open unit balls around $0$ in $X$ and $X/U$, respectively.
It should immediately follow that $\pi$ is an open map and the operator norm of $\pi$ is at most $1$. But can we show that the operator norm is precisely equal to $1$ (unless $X=U$, of course)?
 A: It is also immediate from $\pi(B_1^X(0))=B_1^{X/U}(0)$. If the norm of $\pi$ were strictly less than $1$, we would have $\pi(B_1^X(0))\subseteq \|\pi\|B_1^{X/U}(0)\subsetneq B_1^{X/U}(0)$.
A: Fix $\varepsilon \in (0,1)$. If $S$ is proper (and closed) by Riesz's lemma applied to $1-\varepsilon$ there exists $v \in X$ such that $\|v\| = 1$ and $\|\pi(v)\| = d(v,S) \geq 1-\varepsilon$. Hence $\|\pi\| \geq 1-\varepsilon$; now let $\varepsilon \to 0$.
A: If $U\neq X$, then your projection has norm equal to $1$. Assume it doesn't and let $c=\Vert \pi\Vert<1$, then we have for every $x\in X$
$$ \Vert \pi x\Vert \leq c\Vert x\Vert.$$
This means there exists $u_1\in U$ such that
$$ \Vert x + u_1\Vert \leq (1+c)/2 \Vert x \Vert.$$
Now we apply the same argument to $x+u_1$ and find $u_2\in U$ such that
$$ \Vert x +u_2\Vert \leq [(1+c)/2]^2 \Vert x\Vert.$$
Thus, by induction we find a sequence $(u_n)_{n\geq 1}\subseteq U$ that converges to $x$. As $x$ as arbitrary, we get that $U$ is dense and as $U$ is closed, we get $U=X$.
