Modern Abstract Analysis

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10:02 AM

In the Banach space $C[0,1]$ consider the subspace $M=\lbrace g \in C[0,1]: \int_{0}^{1}g(t)dt=0 \rbrace $ Show that M is closed in $C[0,1]$ and calculate the quotient norm $(\|f+M \|)$ where $f(t)=\sin(\pi t)$ for all $t \in [0,1]$. Probably it's easier than I think but I don't know how do th...

Maybe I am missing something obvious, but the part about finding the norm in the quotient space seems non-trivial to me.

It is basically the question of finding a function $g\in M$ such that $\|f-g\|_\infty = \sup\limits_{t\in[0,1]} |f(t)-g(t)|$ is minimal.

This has a hint of problem from calculus of variations, but there the function we want to optimize is usually an integral, not some kind of supremum.

Maybe I misunderstood something, but it seems that $\sup\limits_{m\in M}\|m\|_\infty=+\infty$ and thus $\inf\limits_{m\in M}(\|\sin\pi t\|-\|m\|)=-\infty$. — Martin Sleziak 8 mins ago

I also think that the norm must less than one. For example, if we take $f(t)=\sin\pi t$ and $g(t)=-\frac16\sin(3\pi t)$ than $\|f(t)-g(t)\|_\infty<1$. Instead of detailed computation I'll add link to WA and also to plot of the graphs. Finding the actual infimum seems to be a more difficult problem. — Martin Sleziak 6 mins ago

Yeah the greater than proof is fishy now.. I will check it again. But norm is a non negative quantity so cannot be $-\infty$ for sure — zenith 26 secs ago

WolframAlpha gives the following:

Yes, that's true. Basically all I am saying is that the way it is currently written, this estimate only gives $\|f\|_{X/M}\ge-\infty$, which is nothing new - as you correctly say. I left a few comments about this question in the functional analysis chatroom - maybe somebody who has an idea how to continue will notice that it was mentioned there. (Of course, you're welcome in that room, too.) — Martin Sleziak 18 secs ago

Feel free to ping me if you have an idea how to continue with the above question.

For instructions how to render MathJax(TeX) in chat see this post on meta or go directly to robjohn's website.

Martin Sleziak

10:31 AM

Maybe it's a bit easier to work with $\sin t$ and $C[0,\pi]$.

The problem is then essential: What is $\inf \|h\|_\infty$ for continuous functions such that $\int_0^\pi h(t) dt =1$.

Where the constant function $h(t)=\frac1\pi$ seems to be the natural choice to get equality.

Martin Sleziak

10:50 AM

Hmm, the inequalities I wrote above are not correct....

The first inequality is in the wrong direction.

Still maybe in some way I should be able to show $\|h\|_\infty\ge\frac1\pi$....?

I am probably just overcomplicating things trying to derive inequalities such as above. (It seems that I wrote all inequality signs in exactly the opposite direction.)

zenith

@MartinSleziak Btw why do you need the integral to be 1. The kernel of the map is with integral zero .. no?

Martin Sleziak

Yes.

So I am looking at $h=f-g$ where $f$ is given and $g\in\ker F$.

Which is equivalent to saying that $F(h)=F(f)$.

So I need $\int_0^\pi h(t) = \int_0^\pi \sin t$ for the modified problem (with $C[0,\pi]$).

Or $\int_0^1 h(t) dt = \int_0^1 \sin\pi t dt$ for the original problem with $C[0,1]$.

Probably changing the interval is not much of a simplification. And I should have written that the integral is equal to 2, not 1; if I work on $[0,\pi]$.

zenith

11:06 AM

So we are trying to find $left\lVert h\right\rVert$_{\infty} where $h \in C[0,\pi]$ and $\int_{0}^{\pi}h(t)dt=2$.. right?

$\left\lVert h\right\rVert_{\infty}$

Martin Sleziak

Yes, we are trying to find $\inf\|h\|_\infty$ where we give such conditions on $h$.

Or, in the original problem we change it to $\int h(t) dt = \frac2\pi$.

@zenith I have just posted an answer, maybe there it is explained a bit clearer.

zenith

Si, I am looking at it.. :)

Martin Sleziak

So I'd guess this might be solved and it was easier than I thought.

Maybe because I was looking at $f(t)=\sin\pi t$, I was always looking at $g\in M$ such that $g(0)=g(1)=0$. After I imposed this additional restriction (by mistake, more-or-less unconsciously), I made the problem more difficult.

zenith

I am thinking .. that to prove the norm by the bi-directional inequalities one needs to guess the value first .. is there a way to do this?

I mean for other f

Martin Sleziak

11:22 AM

Well, in this case the minimum was attained using constant function $h$. Or in other words, the minimum was attained at $g\in M$, where $g=f-C$ where $C$ is constant.

To find a constant function with the prescribed value of integral is not difficult.

Of course, in a different problem (with different subspace $M$), it might happen that the function we are looking for is not a constant function, and it would be more difficult to guess the value of the quotient norm.

BTW I have to admit that I have not worked with quotient normed spaces for some time. Is it the case that the infimum in the definition of the quotient norm is attained at least if the original space is complete (i.e., it's a Banach space)?

Probably not, otherwise we would not need Riesz's lemma, but I cannot come up with a counterexample immediately.

Although I am not really sure whether counterexamples to Riesz's lemma (such as here: A stronger statement of Riesz's lemma) will also give me counterexamples to the claim that quotient norm is attained.

zenith

11:39 AM

Its related to the best approximation to a closed subspace no?

Martin Sleziak

Yes, it deffinitely is.

zenith

I remember studying the best approximation is unique in a Hilbert space .. but we don't need uniqueness

Martin Sleziak

This post on the main seems related to what I asked above: On the norm of a quotient of a Banach space.

zenith

It seems its based on the reflexivity of the subspace

Martin Sleziak

I'd guess if we modify the example we have been working with by changing the subspace $M$ to add some further conditions, we could get a counterexample.

Probably $M'=\{g\in C[0,1]; \int_0^1 g(t)=0, g(0)=g(1)=0\}$ should work.

My feeling is that it is sufficiently similar that we get the same norm, but the boundary condition disallow the use of constant function. (Assuming my answer is correct, that is the only possibility to actually attain the infimum.)

The accepted answer to the question I linked above mentions this:

> 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.

So it seems that this is indeed tied to reflexivity.

Martin Sleziak

12:00 PM

@JyrkiLahtonen I'll just mention that the reason I've added a tag to your post (and bumped it) was that I think that if the tag exists at least the most relevant past question should also get the tag. I hope it is ok.

Jyrki Lahtonen

12:17 PM

No problem @MartinSleziak. Thanks for the maintenance work.

Martin Sleziak

I would add further edits to the question/answer if I saw other things to improve. In this case I bumped it and the only change was the addition of this tag.

I will add that I find the question very interesting and useful. (And also well-written, but this comes as no surprise, considering who is the OP.)

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