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So this is an early exercise in Conway's A Course In Functional Analysis. I'm trying to get to grips with this upto open mapping and closed graph to see if I want to do any more functional analysis. Analysis isn't really my thing but knowing lots of math is empowering and yadda yadda. Anyway the problem:

Let $\mathcal{H}=\{f:[0,1]\rightarrow \mathbb{F} \,:\, f\text{ is absolutely continuous, }f(0)=0, f'\in\mathcal{L}^2(0,1)\}$ (where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$).

Define $\langle f,g\rangle=\int_0^1 f'(t)\overline{g'(t)}\, dt$

Show $\mathcal{H}$ is a Hilbert space.

So I'm happy that $\langle \cdot,\cdot\rangle$ defines an inner product, I just want to show completeness in the metric induced by the norm. So my first stab was given a cauchy sequence $\{f_n\}$ to define $$g(x)=\lim_{n\rightarrow\infty} f_n'(x)\text{ (in $\mathcal{L}^2$'s norm)}$$ This defines $g$ almost everywhere as the sequence $f'(n)$ is Cauchy in $\mathcal{L}^2(0,1)$ (this is direct from the definition of $\mathcal{H}$'s norm). Then I want to define $f(x)=\int_0^x g(t)dt$.

Now I am struggling to show that $f$ is absolutely continuous (it's been a while since real analysis, I could be missing something simple), I'm happy that FTC deals with $f(0)=0$ and $f'\in\mathcal{L}^2(0,1)$ once I've shown this.

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It will be really embarrassing if there is some fact like "the integral of a $\mathcal{L^2}$ function is a.c." –  James Aug 9 '13 at 17:36

2 Answers 2

up vote 3 down vote accepted

This is more or less the continuity of the Lebesgue integral. Since $g \in L^2(0,1) \subset L^1(0,1)$, you have $$\int_E g \, dx \to 0$$ as $\mu(E) \to 0$, where $\mu$ is the Lebesgue measure of $E$.

Now, take $\varepsilon > 0$ arbitrary and pick $\delta > 0$, such that $\int_E g \,dx \le \varepsilon$ for all measurable $E$ with $\mu(E) \le \delta$. This immediately implies the absolute continuity of $f$.

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Inspecting the inner product makes you think that this is all about the derivatives? So let's focus on $f'$ and see how we get $f$ back from it – i.e., by integrating.

Therefore consider this: $L^2(0,1)\subset L^1(0,1)$. Take the map $T$ defined on $L^2(0,1)$ by $$T\phi(x)=\int_0^x\phi(t)\,dt.$$ Surely, this is an isometry of $L^2(0,1)$ onto $\mathcal{H}$, and you're done.

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Is that inclusion true? I believe you if the interval is closed but I'm imagining something with asymptotic behaviour at the end points which $L^1$ says is ok but $L^2$ doesn't. –  James Aug 9 '13 at 18:06
    
Also, as this is in the first chapter I imagine the point is to do this directly. –  James Aug 9 '13 at 18:07
    
@James the inclusion is true. But proving that $T$ maps $L^2(0,1)$ into $H$ still requires you to prove that $T\phi$ is absolutely continuous, so if you don't take this for granted, you still run into the dificulty in your post. –  brom Aug 9 '13 at 18:09
    
The inclusion is true because of the Cauchy–Schwarz inequality since the constant function $1$ is in $L^2(0,1)$. And these spaces don't care about the end points of the interval, since they have measure zero. –  Harald Hanche-Olsen Aug 9 '13 at 19:14

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