Week 13 -- BP2

[Victor Ivrii]

Prove that if $f=\ln |x|$ then
\begin{equation*}
f'(\varphi)= pv \int x^{-1}\varphi (x)\,dx ,
\end{equation*}
where $f'$ is understood in the sense of distributions and the integral is understood as a principal value integral.

[Adam Gao]

$f'(x) = x^{-1}$

$f'(\varphi) = \int f'(x) \varphi(x)\,dx = \int x^{-1}\varphi (x)\,dx = pv\int x^{-1}\varphi (x)\,dx$

Concern/Confusion: Since the derivative of log absolute value of x is same as the derivative of log of x, why do we use a principal value integral instead of a normal integral? Did I somehow abuse notation?

[Victor Ivrii]

Because in $\varphi(0)\ne 0$ even improper $\int x^{-1}\varphi(x)\,dx$ does not exist. Because of this all your arguments are wrong. Not wrong-wrong-wrong, but wrong.

[Adam Gao]

Sorry, I am still confused. Why do we assume $\varphi(0)\ne 0$?

Would I be starting in the right direction with this instead?

$f'(\varphi) = f(\varphi ') = \int f(x) \varphi'(x)\,dx = \int \mathrm{ln}|x| \varphi'(x)\,dx$

[Victor Ivrii]

Sorry, I am still confused. Why do we assume $\varphi(0)\ne 0$?

Would I be starting in the right direction with this instead?

$f'(\varphi) = f(\varphi ') = \int f(x) \varphi'(x)\,dx = \int \mathrm{ln}|x| \varphi'(x)\,dx$

Why do you assume that $\varphi(0)= 0$?? We should try any test functions, including those with $\varphi(0)\ne 0$.

You started in the correct direction but you cannot integrate by parts in $\int_{-\infty}^\infty$. Instead you need to look what vp means

[Adam Gao]

I had a sign error as well. I hope this solution makes more sense.

$f'(\varphi) = -f(\varphi ') = -\int f(x) \varphi'(x)\,dx = -\int \mathrm{ln}|x| \varphi'(x)\,dx = -\int_{-\infty}^{\infty}\mathrm{ln}|x| \varphi'(x)\,dx$

Using limits to evaluate an improper integral

 $= -\lim_{R \to \infty} \int_{-R}^{R}\mathrm{ln}|x| \varphi'(x)\,dx$

Then by integration by parts:

 $= -\lim_{R \to \infty}(\mathrm{ln}|R| \varphi(R) - \mathrm{ln}|-R| \varphi(-R)) + \lim_{R \to \infty}(\int_{-R}^{R}x^{-1}\varphi(x)\,dx) = 0 + \lim_{R \to \infty}\int_{-R}^{R}x^{-1}\varphi(x)\,dx = pv\int x^{-1}\varphi (x)\,dx$

The nonintegral limit term dissappeared because $ln|x|$ is even. I also had to assume $\varphi$ was even to make the proof work. I am not sure if that is true.

[Victor Ivrii]

It is a wrong step in the correct direction. Since $\varphi$ is compactly supported, there is no divergence at infinity, but unless $\varphi(0)=0$, there is divergence at $x=0$.

[Adam Gao]

Wrong step in the correct direction meaning I shouldn't use limits but I should eventually integrate by parts?

Should I use local integration?

[Victor Ivrii]

No, you should use limits. But what does it mean $vp-\int_{-\infty}^\infty f(x)\,dx$ when divergence is not at  infinity but at $a$?

See https://en.wikipedia.org/wiki/Cauchy_principal_value--Just definition

[Adam Gao]

New proof evaluating the limit at $x=0$:

$f'(\varphi) = -f(\varphi ') = -\int f(x) \varphi'(x)\,dx = -\int \mathrm{ln}|x| \varphi'(x)\,dx = -\int_{-\infty}^{\infty}\mathrm{ln}|x| \varphi'(x)\,dx$

Using limits to evaluate an improper integral

 $= -\lim_{\epsilon \to 0^+} \left[ \int_{-\infty}^{-\epsilon}\mathrm{ln}|x| \varphi'(x)\,dx + \int_{\epsilon}^{\infty}\mathrm{ln}|x| \varphi'(x)\,dx \right]$

Then by integration by parts, we have:

$\lim_{\epsilon \to 0^+} \int_{-\infty}^{-\epsilon}\mathrm{ln}|x| \varphi'(x)\,dx$

$ = \lim_{\epsilon \to 0^+} \left [\mathrm{ln}|-\epsilon| \varphi(-\epsilon) - \mathrm{ln}|-\infty|\varphi(-\infty)\right] - \lim_{\epsilon \to 0^+}\int_{-\infty}^{-\epsilon}x^{-1}\varphi(x)\,dx$

and

$\lim_{\epsilon \to 0^+} \int_{\epsilon}^{\infty}\mathrm{ln}|x| \varphi'(x)\,dx$

$ = \lim_{\epsilon \to 0^+} \left[ \mathrm{ln}|\infty| \varphi(\infty) - \mathrm{ln}|\epsilon|\varphi(\epsilon)\right] - \lim_{\epsilon \to 0^+}\int_{\epsilon}^{\infty}x^{-1}\varphi(x)\,dx$

By $\mathrm{ln}|x|$ and $\varphi(x)$ being even (is this correct?) we get:

$\lim_{\epsilon \to 0^+}\left [\mathrm{ln}|-\epsilon| \varphi(-\epsilon) - \mathrm{ln}|-\infty|\varphi(-\infty) + \mathrm{ln}|\infty| \varphi(\infty) - \mathrm{ln}|\epsilon|\varphi(\epsilon)\right] = 0$

so

$f'(\varphi)$

$= -\lim_{\epsilon \to 0^+} \left[ \int_{-\infty}^{-\epsilon}\mathrm{ln}|x| \varphi'(x)\,dx + \int_{\epsilon}^{\infty}\mathrm{ln}|x| \varphi'(x)\,dx \right] = \lim_{\epsilon \to 0^+}\int_{-\infty}^{-\epsilon}x^{-1}\varphi(x)\,dx + \lim_{\epsilon \to 0^+}\int_{\epsilon}^{\infty}x^{-1}\varphi(x)\,dx = pv\int \mathrm{ln}|x| \varphi'(x)\,dx$

[Jingxuan Zhang]

Adam: no you cannot and don't need to assume $\varphi$ is even, but rather you would need to use $\varphi$ is smooth. The gap in your argument is smoothed by
$$(\varphi(-\varepsilon)-\varphi(\varepsilon))\ln\varepsilon\to 0 \text{ as } \varepsilon\downarrow0.$$

[Victor Ivrii]

Yes, we need to assume that $g(x)$ is smooth. Actually, condition is much weaker, just a bit stronger than the continuity of $g$.