Showing that a function is zero a.e.

[AxiomOfChoice]

I've TeX'ed this up directly from my class notes. The argument purports to show that a function $g\in L^2[0,1]$ is zero almost everywhere (a.e.). I dont' see how...can someone help me fill in the missing steps? I'm with him pretty much until the last line...

Let $x_n \to 0$, $x'_n \to g$, both in the $L^2$-norm. (Assume that the $x_n$ are $C^\infty$ with compact support contained inside $(0,1)$.) We want to show $g = 0$ a.e. First, note that $x_n(t) = \int_0^t x_n'(s)ds$ by the FTC. By Holder's inequality, $g\in L^2[0,1]$ implies $g\in L^1[0,1]$:
$$\int_0^1 |g(t)|dt \leq \left( \int_0^1 |g(t)|^2 dt \right)^{1/2} \left( \int_0^1 1^2 \ dt \right)^{1/2} = \left( \int_0^1 |g(t)|^2 dt \right)^{1/2} < \infty.$$
Hence (and he doesn't say so in the notes, but I'm pretty sure this also follows from Holder)
$$\left| \int_0^t (g(s)-x_n'(s))ds \right| \leq \left( \int_0^t |g-x_n'|^2 \right)^{1/2} \left( \int_0^t ds \right)^{1/2} \leq \|g - x_n'\|_{L^2} \cdot 1 \to 0 \quad \text{as }n\to \infty.$$
It follows that $g = 0$ a.e.

Again, the last line of the proof, and what he is trying to do, confuses me, but I think I've provided all the assumptions he established at the onset of this example. However, if something seems amiss, please let me know, and I'll try my best to flesh it out. Thanks!

 

[AxiomOfChoice]

Hmmm...well, I guess I know that
$$\left| \int_0^t (g(s)-x_n'(s))ds \right| = \left| x_n(t) - \int_0^t g(s)ds \right|,$$
so it follows that, for a.e. $t\in [0,1]$, we have
$$\lim_{n\to \infty}\left| \int_0^t (g(s)-x_n'(s))ds \right| = \left| \lim_{n\to \infty} \left(x_n(t) - \int_0^t g(s)ds \right) \right| = \left| \int_0^t g(s) ds \right| = 0.$$
It seems we could conclude $g(s) = 0$ a.e. from THIS, but I don't quite see how...

 

[AxiomOfChoice]

Actually, I'm pretty sure what I just posted is bogus, since we only know $x_n \to 0$ in the $L^2$-norm, not pointwise a.e. So I'm back to being utterly clueless.

 

[micromass]

Hmmm...well, I guess I know that
$$\left| \int_0^t (g(s)-x_n'(s))ds \right| = \left| x_n(t) - \int_0^t g(s)ds \right|,$$

It's ok till here. In what follows:

so it follows that, for a.e. $t\in [0,1]$, we have
$$\lim_{n\to \infty}\left| \int_0^t (g(s)-x_n'(s))ds \right| = \left| \lim_{n\to \infty} \left(x_n(t) - \int_0^t g(s)ds \right) \right| = \left| \int_0^t g(s) ds \right| = 0.$$
It seems we could conclude $g(s) = 0$ a.e. from THIS, but I don't quite see how...

You have used that $\lim_{n\rightarrow +\infty}{x_n(t)}=0$, but you don't know this. You only know convergence in $L^2$-norm, so you can't use convergence a.e.

What you do know is that convergence in $L^2$ implies convergence in measure. And convergence in measure implies that there exists a subsequence that converges to a.e.
So, you cannot use directly that $\lim_{n\rightarrow +\infty}{x_n(t)}=0$, but you DO know it for a subsequence. You can use this to conclude that

$$\int_0^t g(s) ds = 0$$

for all t. Use this to show for every Borel subset A that

$$\int_A g(s)ds=0$$

By choosing [itex]A=\{g>0\}[/tex], one can prove that g=0 a.e.

 

[AxiomOfChoice]

It's ok till here. In what follows:



You have used that $\lim_{n\rightarrow +\infty}{x_n(t)}=0$, but you don't know this. You only know convergence in $L^2$-norm, so you can't use convergence a.e.

What you do know is that convergence in $L^2$ implies convergence in measure. And convergence in measure implies that there exists a subsequence that converges to a.e.
So, you cannot use directly that $\lim_{n\rightarrow +\infty}{x_n(t)}=0$, but you DO know it for a subsequence. You can use this to conclude that

$$\int_0^t g(s) ds = 0$$

for all t. Use this to show for every Borel subset A that

$$\int_A g(s)ds=0$$

By choosing [itex]A=\{g>0\}[/tex], one can prove that g=0 a.e.

WOW...that was a lot for him to have left out. No wonder I couldn't see it. Thanks again, micromass.

Here's another question, though...doesn't this:
$$\int_E |g(t)| \leq \left( \int_E |g(t)|^2 \right)^{1/2} \left( \int_E 1^2 \right)^{1/2} = \sqrt{m(E)} \cdot \left( \int_E |g(t)|^2 dt \right)^{1/2} < \infty$$
show that $L^2(E) \subset L^1(E)$, as long as we restrict ourselves to a measurable set $E$ of finite measure? Or am I missing something?

 

[micromass]

WOW...that was a lot for him to have left out. No wonder I couldn't see it. Thanks again, micromass.

Here's another question, though...doesn't this:
$$\int_E |g(t)| \leq \left( \int_E |g(t)|^2 \right)^{1/2} \left( \int_E 1^2 \right)^{1/2} = \sqrt{m(E)} \cdot \left( \int_E |g(t)|^2 dt \right)^{1/2} < \infty$$
show that $L^2(E) \subset L^1(E)$, as long as we restrict ourselves to a measurable set $E$ of finite measure? Or am I missing something?

That is correct. In general, if $1\leq p\leq q\leq +\infty$ and E is of finite measure, then

$L^q(E)\subseteq L^p(E)$.

Note however that

$$\ell^1\subseteq \ell^2$$

so in this case, the reverse is true.
And, in general there is no inclusion between $L_1(\mathbb{R})$ and $L^2(\mathbb{R})$. So the result above only holds for sets of finite measure!