I've TeX'ed this up directly from my class notes. The argument purports to show that a function [itex]g\in L^2[0,1][/itex] 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...(adsbygoogle = window.adsbygoogle || []).push({});

Let [itex]x_n \to 0[/itex], [itex]x'_n \to g[/itex], both in the [itex]L^2[/itex]-norm. (Assume that the [itex]x_n[/itex] are [itex]C^\infty[/itex] with compact support contained inside [itex](0,1)[/itex].) We want to show [itex]g = 0[/itex] a.e. First, note that [itex]x_n(t) = \int_0^t x_n'(s)ds[/itex] by the FTC. By Holder's inequality, [itex]g\in L^2[0,1][/itex] implies [itex]g\in L^1[0,1][/itex]:

[tex]

\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.

[/tex]

Hence (and he doesn't say so in the notes, but I'm pretty sure this also follows from Holder)

[tex]

\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.

[/tex]

It follows that [itex]g = 0[/itex] 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!

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# Showing that a function is zero a.e.

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