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Let's consider a function $f\in L^2(\mathbb{R})$ for which the second weak derivative exists and lie in $L^2(\mathbb{R})$, i.e. there exists $f''\in L^2(\mathbb{R})$ such that for all $\varphi\in C_0^\infty(\mathbb{R})$ the following integral equation stands: $$ \int\limits_\mathbb{R}f(x)\varphi''(x)dx=\int\limits_\mathbb{R}f''(x)\varphi(x)dx. $$

My question is, having this can we assume that weak $f'$ also exists in $L^2(\mathbb{R})$?

Suppose we found a normal (not generalized) function $g:\mathbb{R}\to\mathbb{C}$ such that for all $\varphi\in C_0^\infty(\mathbb{R})$ $$ \int\limits_\mathbb{R}f(x)\varphi'(x)dx=-\int\limits_\mathbb{R}g(x)\varphi(x)dx. $$ Then $$ \langle -f'', f \rangle_{L^2}=-\int\limits_\mathbb{R}f''(x)f(x)dx=\int\limits_\mathbb{R}g(x)g(x)dx=\|g\|_{L^2}^2 $$ which means, that $g$ is in $L^2(\mathbb{R})$. But what guarantees us the existence of such $g$?

Glinka
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2 Answers2

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Define $ f'(x) = \int_0^x f''(t)d t $, then $|f'(y)|\leq (\int _0^y |f''|^2 )^{1/2}\sqrt{y}\leq ||f''||_2 \sqrt{y}$.

Therefore $$\int\limits_{\mathbb{R}} f''(x) \phi (x) d x = -\int\limits_{\mathbb{R}} f'(x) \phi '(x) d x $$

And so $$\int\limits_{\mathbb{R}} f(x) \phi''(x) d x=-\int\limits_{\mathbb{R}} f'(x) \phi '(x) d x $$

Glinka
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clark
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  • Thank you for your answer! I don't understand how you get the first inequality – Glinka Jun 14 '16 at 14:20
  • @Glinka integration by parts using that $ \lim _{y \rightarrow \pm \infty}f'(y)\phi(y)=0$ (from our inequality), gives $\int (\int_0^x f''(t))' \phi dt = -\int f' \phi ' $ – clark Jun 14 '16 at 14:25
  • I meant, how to derive the inequality itself. How to prove it. Can you add more terms in between? – Glinka Jun 14 '16 at 14:35
  • @Glinka Oh I see, this is from the Cauchy Schwarz for $f''(y)\cdot 1$ – clark Jun 14 '16 at 14:45
  • How does this prove that $f'\in L^2(\mathbb{R})$ – Charuvinda Dec 01 '20 at 16:26
  • @Charuvinda It doesnt. It shows existence. That $f' \in L^2$ is proven in the OP's question, he just wanted existence. – clark Dec 01 '20 at 23:52
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This is clear if you look at the Fourier transform. If $f,f''\in L^2$ then $$\int|\hat f(\xi)|^2<\infty$$ and $$\int|\xi|^4|\hat f(\xi)|^2<\infty,$$and hence $$\int|\xi|^2|\hat f(\xi)|^2<\infty,$$because $|\xi|^2\le\max(1,|\xi|^4)$.

David C. Ullrich
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  • Thank you, that's very elegant! – Glinka Jun 14 '16 at 14:31
  • Why the inverse fourier transform of $(2\pi i \xi)^2\hat{f}(\xi)$ is the object we seek? Why this function $g$ is the weak derivative of $f$? – Glinka Jun 14 '16 at 16:52
  • Actually $g$ is the inverse transform of $2\pi i\xi\hat f(\xi)$. This is basic stuff about weak derivatives versus the Fourier transform. Omitting all the constants, since they vary from book to book depending on how one normalizes the Fourier transform: If $\phi$ is smooth with compact support then an integration by parts shows that $\widehat{\phi'}(\xi)=i\xi\hat\phi(\xi)$. So by Plancherel and the definition of the weak derivative $<g,\phi>=-<f,\phi'>=-<\hat f,i\xi\hat\phi(\xi)>$... – David C. Ullrich Jun 14 '16 at 17:15