What is the proof that M is a convex and closed subset of C[0,1]?

[D_Miller]

I am taking a self-study individual course on convex analysis but I'm having some troubles with the basics as I'm trying to do the exercises in my notes.

I'm asked to consider the space C[0,1] of continuous, complex-valued functions on [0,1], equipped with the supremum norm \|\cdot\|_{\infty}. Let M be the subset of C[0,1] consisting of all those functions f such that

\int_{0}^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^{1}f(x)dx=1.

I now have to show that M is convex and closed in C[0,1]. This is vastly different from anything else I've done in this course so far, and I really have no clue how to start. Any help would be greatly appreciated.

 

[micromass]

What have you tried so far? Can you write out what you have to prove?

 

[D_Miller]

What have you tried so far? Can you write out what you have to prove?

Yeah, if I'm not mistaken I have to prove that for f,g\in M and 0<t<1, the fuction h_t=(1-t)f+tg also lies in M. Then I have to show that the complement of M is open or, equivalently, that the closure of M is equal to M itself.

 

[micromass]

Yes. So for convexity, you'll need to calculate

\int_0^{1/2}{h_t(x)dx}-\int_{1/2}^1{h_t(x)dx}

and show that this equals 1.

 

[D_Miller]

Yes. So for convexity, you'll need to calculate

\int_0^{1/2}{h_t(x)dx}-\int_{1/2}^1{h_t(x)dx}

and show that this equals 1.

Of course. That makes sense and I have now done it. Any idea as to which of my closed set definitions would be easiest to use?

 

[micromass]

Hmm, to me it looks easiest to work with sequences. Take a sequence f_n in M such that f_n\rightarrow f. You'll need to show that f is in M...

 

[D_Miller]

I'm sorry but these kind of proofs always confuse me. I take it I'm supposed to use the result that a set is closed iff it contains all its limit points and then look at

\int_{0}^{\frac{1}{2}}f_t(x)dx-\int_{\frac{1}{2}}^{1}f_t(x)dx=1

and prove it by letting limits pass through the integrals. I just have trouble getting the ball rolling.

 

[micromass]

Well, can you come up with any theorems that allow you to exchange limit and integral? You should have seen those theorems. Remember that the convergence is uniform here...

 

[D_Miller]

Perhaps I've overlooked something, but how can you be sure that the convergence is uniform? I was thinking about the convergence theorems of Lebesgue integration, but if the convergence is uniform, we can interchange the limit and the integral for Riemann integrals as well.

 

[micromass]

Well, if I say f_n\rightarrow f. In what metric space am I working then. More important: which norm am I using?

 

[D_Miller]

Uniform convergence would be in \mathbb{R}, but the norm is stated in the assignment itself to be the sup-norm.

 

[micromass]

Well, we have that a sequence of function converge for the sup-norm if and only if they converge uniformly on \mathbb{R}.

It's not all that hard to prove: just write out the definitions of both convergences and you'll see easily that they're the same...

 

[D_Miller]

Yeah, it stands to reason that a sequence f_n:M\rightarrow \mathbb{R} converges uniformly to f:M\rightarrow \mathbb{R} iff \lim_{n\rightarrow\infty}\| f_n-f\|_S=0.

But is it enough to write that

\int_{0}^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^{1}f(x)dx=\lim_{n\rightarrow \infty}\int_{0}^{\frac{1}{2}}f_ndx-\lim \int_{\frac{1}{2}}^{1}f_ndx=1?

 

[micromass]

Yes, it is enough to write that. You probably seen a theorem that allows you to exchange integrals and limits under uniform convergence. It's that theorem you use...

 

[D_Miller]

Hmm. I do have a bad habit of over-complicating mathematics. Thanks for your help though.

There is a second part to the problem, which I'm not supposed to do. Apparently it's relevant to different course, but just for the heck of it, it asks me to show that M does not contain a vector with the smallest norm.

I assume the result that every nonempty, closed, convex set in a Hilbert space contains a unique element of smallest norm is involved in one way or another. As I understand it, I have to see whether it's possible that \| f\|_{\infty}<1. I have done the calculation that

\int_{0}^{1/2}fdx-\int_{1/2}^{1}fdx\leq \left|\int_{0}^{1/2}fdx\right|+\left|\int_{1/2}^{1}fdx\right| \leq \int_0^1 |f(x)|dx\leq \|f\|_{\infty}

Which (I guess) shows that you can come arbitrarily close but not much else. Any idea as to how this proof could be done correctly? Bear in mind, this is not actual homework, so you don't need to be entirely sure that your proof is correct. I won't hold you responsible anyway ;).

 

[micromass]

Well, your estimation shows us that \|f\|_\infty\geq 1. Now it is possible that the equality holds, but the function that gives you equality is not unique.

Let f be the characteristic function on [0,1/2], then \|f\|_\infty=1.
But let g be the characteristic function on [1/2,1], then \|g\|_\infty=1 to.

So there is no unique function f such that \|f\|_\infty=1.

 

[D_Miller]

Yes, but how does this eliminate the possibility that there are simply multiple vectors with the smallest norm? The task is to show that M doesn't contain any such vectors at all.

 

[D_Miller]

I'm terribly sorry to kick this thread up the board when others probably have more urgent problems, but I'm very curious as to why/if micromass' previous argument (post #16) is enough to prove the stated problem. I'd appreciate it if anyone took the time to elaborate...

 

[Dick]

I'm terribly sorry to kick this thread up the board when others probably have more urgent problems, but I'm very curious as to why/if micromass' previous argument (post #16) is enough to prove the stated problem. I'd appreciate it if anyone took the time to elaborate...

No. micromass' examples f and g aren't in M. They don't satisfy the integral condition. The function h=f-g does and has |h|_infinity=1. And it's essentially the only function with norm=1 that does. But it's not continuous. So it can't be in M.

 

[micromass]

I'm sorry, my post 16 was rubbish.

Basically, you take a function f such that \|f\|_\infty=1.
Then, we can prove with the continuity of f, that f=1 on [0,1/2[ and f=0 on ]1/2,1]. Thus the minimum, if it exists, is unique...

 

[D_Miller]

No. micromass' examples f and g aren't in M. They don't satisfy the integral condition. The function h=f-g does and has |h|_infinity=1. And it's essentially the only function with norm=1 that does. But it's not continuous. So it can't be in M.

I may be missing something completely obvious, but why is this true? Thanks for your time, guys.

 

[Dick]

I may be missing something completely obvious, but why is this true? Thanks for your time, guys.

It is kind of obvious, isn't it? If |f|<=1, then the absolute value of both of the integrals are <=1/2. The only way you can have a-b=1 with |a|,|b|<=1/2 is a=1/2 and b=(-1/2).