# Showing that a function is zero a.e.

1. Aug 17, 2011

### 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!

Last edited: Aug 17, 2011
2. Aug 17, 2011

### 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...

Last edited: Aug 17, 2011
3. Aug 17, 2011

### 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.

4. Aug 17, 2011

### micromass

Staff Emeritus
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 $A=\{g>0\}[/tex], one can prove that g=0 a.e. 5. Aug 17, 2011 ### AxiomOfChoice 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 [itex]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?

6. Aug 17, 2011

### micromass

Staff Emeritus
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!