On the norm of a quotient of a Banach space. Let $E$ be a Banach space and $F$ a closed subspace. It is well known that the quotient space $E/F$ is also a Banach space with respect to the norm 
$$
\left\Vert x+F\right\Vert_{E/F}=\inf\{\left\Vert y\right\Vert_E\mid y\in x+F\}.
$$
Unfortunately in a set of lecture notes on (Lie) group representations (material for our study group) the author accidentally used here $\min$ instead of $\inf$. Probably a mostly harmless booboo, because at that point it was only needed to get a Banach space structure on the quotient, and we will probably be concentrating on Hilbert spaces anyway, where the problem does not arise. 
Namely from Rudin's Functional Analysis I could not find a proof that the minimum should always be attained. Except in the case of a Hilbert space, where an application of parallelogram law (the sum of the squared norms of the two diagonals of a parallelogram equals that of the four sides) allows us to find a Cauchy sequence among a sequence of vectors $(y_n)\subset x+F$ such that 
$$\lim_{n\to\infty}\left\Vert y_n\right\Vert_E=\left\Vert x+F\right\Vert_{E/F}.$$
But anyway, the suspicion was left that the infimum is there for a reason (other than conveniently allowing us to sweep this detail under the rug at that point of the development of theory), so in the interest of serving our study group I had to come up with a specific example, where the minimum is not achieved. It's been 25 years since I really had to exercise the Banach space gland in my brain, so it has shrunk to size of a raisin. Searching this site did help, because I found this question. There we have $E=C([0,1])$, the space of continuous real functions on $[0,1]$ equipped with the sup-norm. If we denote by $\Lambda$ the continuous functional
$$
\Lambda: E\to\mathbb{R},f\mapsto\int_0^{1/2}f-\int_{1/2}^1f
$$
and let $F=\ker\Lambda$, then the answer to the linked question proves that there is no minimum sup-norm function in the coset $\Lambda^{-1}(1)$. 
So I have a (counter)example, and the main question has evolved to:

When can we use minimum in place of infimum in the definition of the quotient space norm?

My thinking:


*

*It seems to me that the answer is affirmative, if $F$ has a complement, i.e. we can write $E=F\oplus F'$ as a direct sum of two closed subspaces such that the norm on $E$ is
equivalent to the sum of the norms on $F$ and $F'$-components.

*But the first point also raises the suspicion that the question may be a bit ill-defined (and uninteresting) in the sense that the answer might depend on the choice of the norm $\left\Vert\cdot\right\Vert_E$ among the set of equivalent norms. However, if we, for example, perturb the sup-norm of $C([0,1])$ in the above example by multiplying the functions with a fixed positive definite function before taking the sup-norm, the argument seems to survive, so may be replacing the norm with an equivalent one is irrelevant?


So to satisfy my curiosity I also welcome "your favorite example" (one with a finite-dimensional $F$ would be nice to see), where we absolutely need the infimum here. Bits about any sufficient or necessary conditions for the minimum to be sufficient or (as a last resort :-) pointers to relevant literature are, of course, also appreciated.
 A: James's theorem asserts the following: if every continuous linear functional on $E$ attains its norm then $E$ is reflexive. This means that on every non-reflexive space we can find a codimension 1 subspace for which the quotient norm needs the infimum in place of a minimum.
In the comments there were already a few links to threads discussing explicit examples, let me list them for the sake of completeness:

*

*Given a point $x$ and a closed subspace $Y$ of a normed space, must the distance from $x$ to $Y$ be achieved by some $y\in Y$? where David Mitra mentions this blog post with a nice discussion.


*Distance minimizers in $L^1$ and $L^{\infty}$ contains more examples and links.

You asked for a finite-dimensional example. This is impossible: for if $F$ is a finite-dimensional subspace of $E$ then we can use compactness of closed and bounded sets in $F$ to ascertain that the infimum is in fact a minimum.
Let $x \in E$ be arbitrary. Choose a sequence $y_{n} \in F$ such that $\lVert x - y_n\rVert \to \lVert x\rVert_{E/F}$. By the reverse triangle inequality $\lVert y_n - x\rVert \geq \lvert \lVert y_n\rVert - \lVert x\rVert\rvert$ we see that $\lVert y_n\rVert \leq 3\lVert x\rVert$ for large enough $n$, so the sequence $(y_n)$ is contained in a bounded subset of $F$. Pass to a convergent subsequence $y_{n_i} \to f \in F$ of $(y_n)$ and observe that $$\lVert x\rVert_{E/F} = \lim \lVert x - y_{n_i}\rVert = \lVert x - f\rVert.$$

With a little more work and refining the above idea, one can prove that a sufficient condition is reflexivity of the subspace $F$.
To see this, recall the following facts:

*

*A norm-closed convex subset $C$ of a Banach space is weakly closed.
By the Hahn-Banach separation theorem we can write $C$ as the intersection of closed half-spaces defined by linear functionals, and these half-spaces are weakly closed since the weak topology is the initial topology induced by the linear functionals.


*The closed unit ball of a reflexive Banach space is weakly compact (and hence every closed ball is weakly compact).
Since $E$ is reflexive, it is the dual space of $E^{\ast}$ and the weak and weak*-topologies on $E$ coincide. Since the closed unit ball in a dual space is weak*-compact by Alaoglu's theorem, it is therefore weakly compact.


*Combining 1. and 2. we see that every closed and bounded convex set in a reflexive space is weakly compact. [One can also prove that a closed and bounded convex set in a reflexive space is weakly sequentially compact but we won't need this here].


*The norm on a Banach space is weakly lower semicontinuous: if a net $x_i$ converges weakly to $x$ then $\lVert x \rVert \leq \liminf_i \lVert x_i\rVert$.
By Hahn-Banach there is a norm 1-functional $\varphi \in E^\ast$ such that $\varphi(x) = \lVert x\rVert$. But then $\lVert x\rVert = \lim_{i} \lvert \varphi(x_i)\rvert \leq  \liminf_i \lVert \varphi\rVert\lVert x_i\rVert = \liminf_i \lVert x_i\rVert$.


*The restriction of weak topology on $E$ to a closed subspace $F$ of $E$ is the same as the weak topology of $F$ as a Banach space.
This follows from Hahn-Banach.
We are finally ready to prove the announced result:

Let $F$ be a reflexive subspace of the Banach space $E$. Let $x \in E$ be arbitrary. Then there is $f \in F$ such that $\lVert x\rVert_{E/F}  = \lVert x - f\rVert = \inf_{y \in F} \lVert x - y\rVert$.

Choose a sequence $y_{n} \in F$ such that $\lVert x - y_n\rVert \to \lVert x\rVert_{E/F}$. Again the reverse triangle inequality shows that $\lVert y_n\rVert \leq 3\lVert x\rVert$ for large enough $n$, so the sequence $(y_n)$ is contained in a weakly compact convex subset of $F$. Pass to a weakly convergent subnet $y_{n_i} \to f \in F$ of $(y_n)$ and observe that $$\lVert x\rVert_{E/F} \leq \lVert x -f\rVert \leq \liminf \lVert x - y_{n_i}\rVert = \liminf \lVert x - y_n\rVert = \lVert x\rVert_{E/F}.$$
