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Let $a>0$ and $$\langle f_a,\psi\rangle=\int_{|x|>a} \frac{\psi(x)}{|x|} dx + \int_{|x|<a} \frac{\psi(x)-\psi(0)}{|x|}dx$$ Prove that $f_a$ does not depend on a.

Proof: To prove that $f_a$ does not depend on a we have to take the derivate of the $f_a$ and set it equal to 0.

$$\langle f_a',\psi \rangle =-\langle f_a,\psi'\rangle=-[\int_{|x|>a} \frac{\psi'(x)}{|x|} dx + \int_{|x|<a} \frac{\psi'(x)-\psi'(0)}{|x|}dx] =-[\int_{|x|>a} \frac{\psi'(x)}{|x|} dx + \int_{|x|<a} \psi'(c)dx] =-[\int_{-\infty}^{-a} \frac{\psi'(x)}{|x|} dx + \int_a^{\infty} \frac{\psi'(x)}{|x|} dx + \int_{|x|<a} \psi'(c)dx] =-[\int_{-\infty}^{-a} \frac{\psi'(x)}{|x|} dx + \int_a^{\infty} \frac{\psi'(x)}{|x|} dx + \int_{|x|<a} \psi'(c)dx] =-[\int_{-\infty}^{-a} \frac{\psi'(x)}{|x|} dx + \int_a^{\infty} \frac{\psi'(x)}{|x|} dx + 2a\psi'(c)$$

How do I proceed?

Ivo Terek
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Username Unknown
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    A number of things don't make sense here. First there's no dependence on $f$ on the right hand side, just $a$. Next, you seem to be assuming that you have taken a derivative in $a$, yet in your inner product it's really a derivative in $x$, as in $f_a(x),\psi(x)$ and invoked an integration by parts procedure. – Alex R. Jul 09 '14 at 03:00
  • @AlexR. Can you help me correct the work I have? – Username Unknown Jul 09 '14 at 03:08
  • @AlexR. that is not an inner product, and $f_a$ is a distribution thus inherently the r.h.s. is $f_a$ – DanZimm Jul 09 '14 at 08:26

1 Answers1

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Let $a>0$. Put

$$T_a= \int_{|x| \geq a} \frac{\psi(x)}{|x|} dx + \int_{|x|<a} \frac{\psi(x)-\psi(0)}{|x|}dx$$ and for $\varepsilon\in ]0,a[$: $$T_a(\varepsilon)=\int_{|x| \geq a} \frac{\psi(x)}{|x|} dx + \int_{\varepsilon<|x|<a} \frac{\psi(x)-\psi(0)}{|x|}dx$$

We have that $T_a(\varepsilon)\to T_a$ if $\varepsilon\to 0$.

Now:

$$T_a(\varepsilon)=\int_{|x| \geq a} \frac{\psi(x)}{|x|} dx +\int_{\varepsilon<|x|<a} \frac{\psi(x)}{|x|}dx-\psi(0) \int_{\varepsilon<|x|<a} \frac{1}{|x|}dx$$

We have:

$$ \int_{\varepsilon<|x|<a} \frac{1}{|x|}dx=2\log(\frac{a}{\varepsilon})$$ Hence

$$T_a(\varepsilon)=\int_{|x| >\varepsilon} \frac{\psi(x)}{|x|} dx-2\psi(0)\log(\frac{a}{\varepsilon})$$

We have also for $b>a$:

$$T_b(\varepsilon)=\int_{|x| >\varepsilon} \frac{\psi(x)}{|x|} dx-2\psi(0)\log(\frac{b}{\varepsilon})$$

Hence: $$T_b(\varepsilon)-T_a(\varepsilon)=-2\psi(0)(\log b - \log a)$$ And now if $\varepsilon \to 0$: $$T_b-T_a=-2\psi(0)(\log b-\log a)$$ (So this is $T_a+2\psi(0)\log a$ that is independent of $a>0$).

Kelenner
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