Prove $\lim_{n\to\infty}\frac{\int_{-1}^{1}{f(x)(1-x^2)^n}dx}{\int_{-1}^{1}{(1-x^2)^n}dx}=f(0)$

Solution 1:

Here is a standard argument dealing with this type of problem, often known as approximation-to-the-identity. Here is a slightly general claim, which does not harm the essence of the argument.

Proposition. Let $K_n \in C([-1,1])$ and assume that $(K_n)$ satisfy the following assumtions:

1. Nonnegativity. $K_n(x) \geq 0$ for all $x \in [-1, 1]$ and $n \geq 1$.
2. Normalization. $\int_{-1}^{1} K_n(x) \, \mathrm{d}x = 1$ for all $n \geq 1$.
3. Concentration. For any $\delta > 0$, $\lim_{n\to\infty} \int_{|x|>\delta} K_n(x) \, \mathrm{d}x = 0$.

Then for any $f \in C([-1, 1])$, we have $$ \lim_{n\to\infty} \int_{-1}^{1} f(x)K_n(x) \, \mathrm{d}x = f(0). $$

In other words, $K_n(x)$ approximates the Dirac delta $\delta(x)$, which is "the identity" of the convolution (hence the name approximation-to-the-identity). Before the proof, let us check that this can be applicable to OP's problem. Indeed, set

$$K_n(x) = \frac{(1-x^2)^n}{\int_{-1}^{1} (1-t^2)^n \, \mathrm{d}t}. $$

Then the conditions 1 and 2 are obvious. For the concentration, notice that $(1 - t^2)^n \geq (1 - |t|)^n$, and so,

$$ \int_{|x|>\delta} K_n(x) \, \mathrm{d}t \leq \frac{(1 - \delta^2)^n}{\int_{0}^{1}(1 - t)^n \, \mathrm{d}t} = (n+1)(1 - \delta^2)^n \xrightarrow[n\to\infty]{} 0. $$

Therefore $\int_{-1}^{1} f(x)K_n(x) \, \mathrm{d}x \to f(0)$ by the conclusion of the proposition.

Proof of Proposition. For any $\epsilon > 0$, find $\delta > 0$ so that $|x| < \delta$ implies $|f(x) - f(0)| < \epsilon$. Also, choose a bound $M > 0$ of $f$. Then

\begin{align*} \left| \int_{-1}^{1} f(x)K_n(x) \, \mathrm{d}x - f(0) \right| &\leq \int_{-1}^{1} |f(x) - f(0)|K_n(x) \, \mathrm{d}x \\ &= \int_{|x|<\delta} |f(x) - f(0)|K_n(x) \, \mathrm{d}x + \int_{|x|\geq\delta} |f(x) - f(0)|K_n(x) \, \mathrm{d}x \\ &\leq \int_{-1}^{1} \epsilon K_n(x) \, \mathrm{d}x + \int_{|x|\geq\delta} 2M K_n(x) \, \mathrm{d}x \\ &= \epsilon + 2M \int_{|x|\geq\delta} K_n(x) \, \mathrm{d}x. \end{align*}

So we have

$$ \limsup_{n\to\infty} \left| \int_{-1}^{1} f(x)K_n(x) \, \mathrm{d}x - f(0) \right| \leq \epsilon. $$

But since the left-hand side a fixed number independent of $\epsilon$ and the right-hand side can be made arbitrarily small, letting $\epsilon \downarrow 0$ proves the desired claim. $\square$

Solution 2:

Let $x=y/\sqrt n.$ The expression becomes

$$\frac{\int_{-\sqrt n}^{\sqrt n}f(y/\sqrt n)(1-y^2/n)^n\,dy}{\int_{-\sqrt n}^{\sqrt n}(1-y^2/n)^n\,dy}.$$

Now $(1-y^2/\sqrt n)^n\le e^{-y^2}$ on $[-\sqrt n,\sqrt n].$ And $f$ is bounded. Also note that for fixed $y,$ $f(y/\sqrt n) \to f(0)$ by the continuity of $f$ at $0.$ It follows from the DCT that the limit is

$$\frac{\int_{-\infty}^{\infty} f(0)e^{-y^2}\, dy}{\int_{-\infty}^{\infty} e^{-y^2}\, dy} = f(0).$$