Jtdylktuy

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
$$
leftVert x+FrightVert_{E/F}=inf{leftVert yrightVert_Emid yin x+F}.
$$
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
$$lim_{ntoinfty}leftVert y_nrightVert_E=leftVert x+FrightVert_{E/F}.$$

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
$$
Lambda: Etomathbb{R},fmapstoint_0^{1/2}f-int_{1/2}^1f
$$
and let $F=kerLambda$, 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:

1. It seems to me that the answer is affirmative, if $F$ has a complement, i.e. we can write $E=Foplus F'$ 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.


2. 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 $leftVertcdotrightVert_E$ 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.

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 $yin Y$? 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 $lVert x - y_nrVert to lVert xrVert_{E/F}$. By the reverse triangle inequality $lVert y_n - xrVert geq lvert lVert y_nrVert - lVert xrVertrvert$ we see that $lVert y_nrVert leq 3lVert xrVert$ 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 $$lVert xrVert_{E/F} = lim lVert x - y_{n_i}rVert = lVert x - frVert.$$

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:

1. 
   

   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.

   
   


2. 
   

   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.

   
   



3. 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].



4. 
   

   The norm on a Banach space is weakly lower semicontinuous: if a net $x_i$ converges weakly to $x$ then $lVert x rVert leq liminf_i lVert x_irVert$.

   
   
   

   By Hahn-Banach there is a norm 1-functional $varphi in E^ast$ such that $varphi(x) = lVert xrVert$. But then $lVert xrVert = lim_{i} lvert varphi(x_i)rvert leq liminf_i lVert varphirVertlVert x_irVert = liminf_i lVert x_irVert$.

   
   


5. 
   

   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 $lVert xrVert_{E/F} = lVert x - frVert = inf_{y in F} lVert x - yrVert$.

Choose a sequence $y_{n} in F$ such that $lVert x - y_nrVert to lVert xrVert_{E/F}$. Again the reverse triangle inequality shows that $lVert y_nrVert leq 3lVert xrVert$ 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 $$lVert xrVert_{E/F} leq lVert x -frVert leq liminf lVert x - y_{n_i}rVert = liminf lVert x - y_nrVert = lVert xrVert_{E/F}.$$