Does pointwise convergence imply convergence of integral on C[0,1]?

[ChemEng1]

Homework Statement

Does pointwise convergence imply convergence of integral on C[0,1]?

Homework Equations

The Attempt at a Solution

Would like to verify that f(t)=t^n is a counter example of this. Pointwise convergence goes to 0 on [0,1) and 1 on [1]. Norm(1) is unbounded.

I hope that right. It looks so small written out, but took almost 2 hours putting it together.

Thanks in advance.

 

[jbunniii]

What norm are you using? Is it

\int_{0}^{1}|f(t)|dt?

If so, I disagree that it is unbounded! (In fact, it's bounded by 1 for all n.) Can you please show your calculation?

 

[ChemEng1]

And then I just realized that doesn't work either. Sigh. Norm(1)=1/(n+1) which isn't unbounded. Help. Please?

 

[ChemEng1]

What norm are you using? Is it

\int_{0}^{1}|f(t)|dt?

If so, I disagree that it is unbounded! Can you please show your calculation?

I realized that at the same time you posted it. Sadness.

 

[jbunniii]

Think about a sequence of functions that get arbitrarily tall and narrow as n increases.

 

[ChemEng1]

My intuition is that pointwise convergence does not imply convergence of an integral. Integrals work across an interval and not on points like pointwise convergence would look at. Just not sure how to find an example showing that.

 

[jbunniii]

OK, think about a sequence of functions, where each function is shaped like a triangle. Define the base and height such that the base shrinks to 0 and the height grows to infinity as n goes to infinity. Can you do this in such a way that the area stays constant regardless of n?

Then answer these questions: are these functions in C[0,1]? Do they converge pointwise to a function in C[0,1]? Do they converge to that function (or any function) in the integral norm?

 

[ChemEng1]

How about a function the looks like a right triangle with base 1/n and height n^2 at t=0? The integral norm would always be n/2 which would go to infinity with n. Would pointwise convergence go to 0 as well? (I am a bit unconfident of saying this because of what would happen with the function at t=0 though. For t=(0,1], I would agree.)

 

[Ray Vickson]

Homework Statement

Does pointwise convergence imply convergence of integral on C[0,1]?

Homework Equations

The Attempt at a Solution

Would like to verify that f(t)=t^n is a counter example of this. Pointwise convergence goes to 0 on [0,1) and 1 on [1]. Norm(1) is unbounded.

I hope that right. It looks so small written out, but took almost 2 hours putting it together.

Thanks in advance.

There is a bit of ambiguity in your question. Say you have a sequence {f_n} in C[0,1] that converges pointwise. Do you want the limit function f to also be in C[0,1], or do you allow it to be discontinuous and/or unbounded? (I suspect you mean the latter, but that is for you to clarify.)

RGV

 

[ChemEng1]

I think I got it. Right triangle with base of 1/n and height of n set at t=0. Integral norm goes to 1/2. But function pointwise diverges because of behavior at t=0.

Look good?

 

[jbunniii]

I think I got it. Right triangle with base of 1/n and height of n set at t=0. Integral norm goes to 1/2. But function pointwise diverges because of behavior at t=0.

Look good?

But this isn't a counterexample. The function diverges pointwise, and it also doesn't converge to anything in the integral norm. In order for the latter to occur, there would have to be a function g such that

\int_{0}^{1}|f_n(t) - g(t)| dt \rightarrow 0

as n \rightarrow \infty. All you have is a sequence of functions which all have norm = 1/2. That doesn't imply any kind of convergence.

However, you can modify the sequence so that it will converge pointwise. To do so, try arranging all the triangles so that the left edge of the base starts at t = 0, instead of centering the triangles at t = 0. I claim that this sequence converges pointwise. (To what?)

 

[ChemEng1]

But this isn't a counterexample. The function diverges pointwise, and it also doesn't converge to anything in the integral norm. In order for the latter to occur, there would have to be a function g such that

\int_{0}^{1}|f_n(t) - g(t)| dt \rightarrow 0

as n \rightarrow \infty. All you have is a sequence of functions which all have norm = 1/2. That doesn't imply any kind of convergence.

However, you can modify the sequence so that it will converge pointwise. To do so, try arranging all the triangles so that the left edge of the base starts at t = 0, instead of centering the triangles at t = 0. I claim that this sequence converges pointwise. (To what?)

So construct a function that is a sequence of isosceles triangles with base length 1/n and height n. This function converges to 0 pointwise, right?

Now I need to show that \int_{0}^{1}|f_n(t) - 0| dt \neq 0, correct? (This integral would go to 1, which isn't 0. So I think I'm done.

I see where my mistake was earlier. I showed that it did not converge pointwise, but still needed to determine a function it converges to under integral norm to test to show it goes to 0.

 

[jbunniii]

So construct a function that is a sequence of isosceles triangles with base length 1/n and height n. This function converges to 0 pointwise, right?

Correct, assuming you made the modification I described earlier, so that the triangles are centered at, say, 1/n instead of at 0.

Now I need to show that \int_{0}^{1}|f_n(t) - 0| dt \neq 0, correct? (This integral would go to 1, which isn't 0. So I think I'm done.

This shows that the sequence does not converge to 0 in the integral norm. On the face of it, there is still the possibility that the sequence could converge to some other function in the integral norm. Can you prove that this is not the case?

P.S. I assume you meant to take the limit:

\lim_{n \rightarrow \infty} \int_{0}^{1}|f_n(t) - 0| dt \neq 0