Uniform convergence, mean convergence, mean-square convergence

[Lajka]

Hi,

I was always troubled by the relationships between these modes of convergence (L^1, L^2, and L^{\infty} convergences, to be precise), so I took some books and decided to establish some relations between them. For some, I succeeded, for others I did not. Here's what I did so far:

If I is a finite interval:
L^{\infty} implies L^{1} (proven)
L^{\infty} implies L^{2} (proven)
L^{2} implies L^{1} (proven)

none of the converses are true (found counter-examples for each)

If I is an infinite interval:
L^{\infty} doesn't imply L^{1} (counter-example)
L^{\infty} doesn't imply L^{2} (counter-example)
L^{2} doesn't imply L^{1} (counter-example)

However, I'm yet to see if, for infinite intervals:
L^{1} implies L^{\infty}
L^{2} implies L^{\infty}
L^{1} implies L^{2}

Intuitively, I believe that L^{1} (or L^{2}) cannot imply uniform convergence (I just imagine a Gaussian which shrinks (keeping his height constant)). However, I cannot think of an example to disprove if L^{1} implies L^{2}.

Also, I should say that I don't know how to prove any of this conclusions for infinite intervals, so I'm just trying to find counter-examples (for finite intervals, I just used Cauchy-Schwarz inequality and mean value theorem, but none of that can be used here).

So, any help is appreciated, thanks.

P.S. Just one more thing that caught up in my mind whilst I was writing this: If I take a Gaussian and make new functions by letting it shrink (but increasing its height this time), that sequence would converge to what? I know that's one of the ways to define a delta function, but in a light of this discussion, I would probably say it would converge pointwise to f(x) \equiv 0.

 

[micromass]

Hi Lajka!

Hi,

I was always troubled by the relationships between these modes of convergence (L^1, L^2, and L^{\infty} convergences, to be precise), so I took some books and decided to establish some relations between them. For some, I succeeded, for others I did not. Here's what I did so far:

If I is a finite interval:
L^{\infty} implies L^{1} (proven)
L^{\infty} implies L^{2} (proven)
L^{2} implies L^{1} (proven)

none of the converses are true (found counter-examples for each)

If I is an infinite interval:
L^{\infty} doesn't imply L^{1} (counter-example)
L^{\infty} doesn't imply L^{2} (counter-example)
L^{2} doesn't imply L^{1} (counter-example)

However, I'm yet to see if, for infinite intervals:
L^{1} implies L^{\infty}
L^{2} implies L^{\infty}
L^{1} implies L^{2}

Intuitively, I believe that L^{1} (or L^{2}) cannot imply uniform convergence (I just imagine a Gaussian which shrinks (keeping his height constant)).

For the first two, consider this:

f_n:[0,1]\rightarrow \mathbb{R}:t\rightarrow \left\{\begin{array}{ccc}<br /> 0 &amp; \text{if} &amp; 0\leq t\leq \frac{1}{2}\\<br /> nt-\frac{1}{2}n &amp; \text{if} &amp; \frac{1}{2}\leq t\leq \frac{1}{2}+\frac{1}{n}\\<br /> 1 &amp; \text{if} &amp; \frac{1}{2}+\frac{1}{n}\leq t\leq 1\\<br /> \end{array}\right.

[/QUOTE]
However, I cannot think of an example to disprove if L^{1} implies L^{2}.

Try the function

f_n:[0,1]\rightarrow \mathbb{R}:t\rightarrow \left\{\begin{array}{ccc}<br /> 0 &amp; \text{if} &amp; 0\leq t\leq \frac{1}{n}\\<br /> t^{-2/3} &amp; \text{if} &amp; \frac{1}{n}\leq t\leq 1\\<br /> \end{array}\right.

Also, I should say that I don't know how to prove any of this conclusions for infinite intervals, so I'm just trying to find counter-examples (for finite intervals, I just used Cauchy-Schwarz inequality and mean value theorem, but none of that can be used here).

So, any help is appreciated, thanks.

P.S. Just one more thing that caught up in my mind whilst I was writing this: If I take a Gaussian and make new functions by letting it shrink (but increasing its height this time), that sequence would converge to what? I know that's one of the ways to define a delta function, but in a light of this discussion, I would probably say it would converge pointwise to f(x) \equiv 0.

It doesn't converge pointswise, since it doesn't converge for x=0. It does converge to f(x)=0 almost everywhere. But that's the best you can do.
If you allow distributions, then the functions converge to the Dirac Delta distribution.

 

[Lajka]

Hey micromass! :)

First of all, thanks for resolving my dilemma about the delta function. :)

Now, as for your examples, I only have one problem, and that's that they're defined over finite intervals, and I needed examples of the functions which converge (or not) over infinite intervals. Or maybe I'm missing something here?

For the first two, consider this:

f_n:[0,1]\rightarrow \mathbb{R}:t\rightarrow \left\{\begin{array}{ccc}<br /> 0 &amp; \text{if} &amp; 0\leq t\leq \frac{1}{2}\\<br /> nt-\frac{1}{2}n &amp; \text{if} &amp; \frac{1}{2}\leq t\leq \frac{1}{2}+\frac{1}{n}\\<br /> 1 &amp; \text{if} &amp; \frac{1}{2}+\frac{1}{n}\leq t\leq 1\\<br /> \end{array}\right.

This converges to the Heaviside function, right? great example, but the fact that it's over finite domain bothers me. However, while reading your examples, I thought of this. Observe this function
f_n:\mathbb{R}\rightarrow \mathbb{R}:t\rightarrow \left\{\begin{array}{ccc}<br /> -1 &amp; \text{if} &amp; - \infty &lt; t &lt; - \frac{1}{n}\\<br /> nt &amp; \text{if} &amp; -\frac{1}{n}\leq t\leq \frac{1}{n}\\<br /> 1 &amp; \text{if} &amp; \frac{1}{n} &lt; t &lt; + \infty\\<br /> \end{array}\right.
which also converges to the unit step function, if I'm not mistaken, and it also converges in L^1 and in L^2, but not it L^{\infty}. What do you think, is this an okay example, does it make sense?

Try the function

f_n:[0,1]\rightarrow \mathbb{R}:t\rightarrow \left\{\begin{array}{ccc}<br /> 0 &amp; \text{if} &amp; 0\leq t\leq \frac{1}{n}\\<br /> t^{-2/3} &amp; \text{if} &amp; \frac{1}{n}\leq t\leq 1\\<br /> \end{array}\right.

If not for the fact that it's defined over finite domain, this would be an awesome example. I mean, it still is :D it's just that I needed it to be defined over infinite interval.

Thanks again for your help micromass, and If my reasoning isn't proper here, please let me know!