The challenge is: if $f:\mathbb{R}\to\mathbb{R}$ is a continuous function, then $$\lim_{t\to0+}\frac{1}{\pi}\int_{-1}^1\frac{tf(x)}{t^2+(x-a)^2}\,\mathrm{d}x=f(a)$$ for every $a\in[-1,1]$

My attemps:

  1. Under certainly conditions, i put the limit symbol under the integral sign, but the limit would be $0$. So it is a bad idea.

  2. Therefore, i think use Weierstrass theorem for $f$ in $[-1,1]$, but i can't solve the problem in the case of $f(x)=x^n$.

What do you recommendme?

  • 2
    $\begingroup$ It could be insightful to start proving the result when $f(x)$ is a constant function. In any case, that $1/ \pi$ makes me think we need Fourier series. $\endgroup$
    – Crostul
    Mar 2 at 22:14
  • $\begingroup$ To add to @Crostul's suggestion, another possibly useful thing could be to assume that $f$ is periodic with a period of $2$ since the equation is only concerned with $a \in [-1, 1]$. (However, this may lead to $f$ being discontinuous at odd integers.) $\endgroup$ Mar 2 at 22:18

I consider the case $a=0$.

Note that $$\lim_{t \to 0 } \int_{-1}^1 \frac{tf(0)}{t^2+x^2} dx = f(0) \lim_{t \to 0} \left( \arctan(1/t) - \arctan(-1/t) \right) = \pi f(0).$$

This essentially reduces the problem to the case where $f(0)=0$.

If $f(0)=0$, by continuity for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|f(x)| \le \epsilon$ for $|x| \le \delta$. It then follows that $$\lim_{t \to 0} \int_{|x| > \delta} \frac{tf(x)}{t^2+x^2} dx = \int_{|x| > \delta} \lim_{t \to0} \frac{tf(x)}{t^2+x^2} dx = \int_{|x|>\delta} 0 \, dx = 0.$$ On the other hand, we have $$ \left| \lim_{t \to 0} \int_{-\delta}^\delta \frac{tf(x)}{t^2+x^2} dx \right| \le \lim_{t \to 0} \int_{-\delta}^\delta \left| \frac{tf(x)}{t^2+x^2} \right| dx \le 2\epsilon \lim_{t \to 0} \int_{0}^\delta \frac{t}{t^2+x^2} dx = \lim_{t \to 0} 2\epsilon \arctan(\delta/t) = 2\pi \epsilon. $$ It follows that $$\left| \lim_{t \to 0} \int_{-1}^1 \frac{tf(x)}{t^2+x^2} dx \right| \le 2\pi \epsilon$$ for all $\epsilon > 0$, so $\lim_{t \to 0} \int_{-1}^1 \frac{tf(x)}{t^2+x^2} dx = 0$.

  • $\begingroup$ At first I didn't understand how your reduction to the $f(0)=0$ case worked, so to clarify it here: we write $f(x)=f(0)+g(x)$, so that $g(0)=0$, and integrate the resulting two terms separately. $\endgroup$
    – Karl
    Mar 2 at 23:22

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