Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

  1. 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.
  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 $\left\Vert\cdot\right\Vert_E$ 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.

share|improve this question
1  
I seem to recall that the minimum is attained in any reflexive Banach space, but I may be mistaken. –  JSchlather Feb 10 '13 at 21:58
1  
I vaguely recall this has to do with the proximinality (or proximality, not sure) of the subspace. But I can't find any good reference. –  1015 Feb 10 '13 at 22:11
2  
See this post. The link in my comment there gives a general method for constructing a counterexample in a nonreflexive space, as well as a concrete example for $\ell_1$. –  David Mitra Feb 10 '13 at 22:48
1  
Example 71 of these notes gives a coherent proof that in any reflexive space, the distance from a point to a closed, convex, non-empty set is attained. Of course, from the other link I mentioned, it follows that the converse holds. –  David Mitra Feb 10 '13 at 23:03
1  
I finally found something: web.cs.dal.ca/~jborwein/cheb.pdf There is a proposition there that states the characterization of reflexive spaces mentioned by David Mitra. Proximal and proximinal are both used: a set $S$ in $E$ is called proximal if $d(x,S)$ is attained for every $x\in E$. –  1015 Feb 10 '13 at 23:09
show 3 more comments

1 Answer

up vote 11 down vote accepted

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:


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:

  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 $\lVert x \rVert \leq \liminf_i \lVert x_i\rVert$.

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

  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 $\lVert x\rVert_{E/F} = \lVert x - f\rVert = \inf_{y \in F} \lVert x - y\rVert$.

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 $$\lVert x\rVert_{E/F} \leq \lVert x -f\rVert \leq \liminf \lVert x - y_{n_i}\rVert = \liminf \lVert x - y_n\rVert = \lVert x\rVert_{E/F}.$$

share|improve this answer
    
Thanks, Martin. This argument should be accessible to me. I will need to find the time to absorb the details before I accept, but looks good. –  Jyrki Lahtonen Feb 11 '13 at 6:17
    
Thanks. Somehow I'm unhappy with the longish and clumsy argument since the facts I outline seem a bit redundant. The basic idea is to find a substitute for norm compactness yielding the minimizer in the finite-dimensional case and weak compactness is the only one I could think of. Probably there are more efficient and clearer ways to put it. Basically the two ideas are: 1) the function $y \mapsto \lVert x-y\rVert$ attains its minimum on every weakly compact convex subset. 2) a minimizing sequence in the reflexive subspace $F$ must be contained in a weakly compact convex subset of $F$. –  Martin Feb 11 '13 at 10:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.