Toronto Math Forum
APM3462018S => APM346––Home Assignments => Web Bonus Problems => Topic started by: Victor Ivrii on April 04, 2018, 12:28:13 PM

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.

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

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 wrongwrongwrong, but wrong.

Sorry, I am still confused. Why do we assume $\varphi(0)\ne 0$?
Would I be starting in the right direction with this instead?
$f'(\varphi) = f(\varphi ') = \int f(x) \varphi'(x)\,dx = \int \mathrm{ln}x \varphi'(x)\,dx$

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

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 $lnx$ is even. I also had to assume $\varphi$ was even to make the proof work. I am not sure if that is true.

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 $x=0$.

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

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 (https://en.wikipedia.org/wiki/Cauchy_principal_value)Just definition

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

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
$$(\varphi(\varepsilon)\varphi(\varepsilon))\ln\varepsilon\to 0 \text{ as } \varepsilon\downarrow0.$$

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