Justify the interchange of limits for $\displaystyle\lim_{n \to \infty} \lim_{m \to \infty} (n+1) \int_0^1 x^nP_m(x)dx$ where $P_m(x)$ is a polynomial

Question

I am trying to prove the following problem:

Let $f(x)$ be a real and continuous function on $[0,1]$. Show that $$\displaystyle\lim_{n \to \infty} (n+1) \int_0^1 x^n f(x) dx = f(1).$$

I have shown that the result holds true if $f(x)$ is a polynomial. By Weierstrass approximation theorem, there exists functions $P_m(x)$ of degree $m$ that converges uniformly to $f(x)$ on $[0,1].$ Because $x^n$ is bounded on $[0,1]$ for all $n$, we know that $x^nP_m(x)$ converges uniformly to $x^nf(x)$ as $m$ approaches infinity. Thus we are reduced to showing that we are able to interchange the limits in

$$\lim_{n \to \infty} \lim_{m \to \infty} (n+1) \int_0^1 x^nP_m(x)dx = \lim_{m \to \infty} \lim_{n \to \infty} (n+1) \int_0^1 x^nP_m(x)dx.$$

I am not sure how to justify this. If you have any other way of solving the problem you are welcome to share. Thank you!

Answer 1

Hint: $$|(n+1) \int_0^{1}x^{n} P_m(x)dx-(n+1) \int_0^{1}x^{n} f(x)dx|$$ $$ \leq (n+1) \int_0^{1}x^{n}|f(x)- P_m(x)| dx$$ $$<\epsilon (n+1)\int_0^{1}x^{n}dx=\epsilon$$ for all $n$ if $m$ is sufficiently large.

Answer 2

Hint for an alternate solution:

Use the substitution $t = x^{n+1}$

$$I = \lim_{n\to\infty}\int_0^1f\left(t^{\frac{1}{n+1}}\right)dt$$

Then use the continuity of $f$ on a closed interval to justify the limit interchange.

Answer 3

What about proving it directly? Let $\delta >0$. Then $$\Big\vert(n+1)\int_0^{1-\delta} x^n f(x) dx\Big\vert \leq \sup_{x\in [0,1]} \vert f(x) \vert \int_0^{1-\delta} (n+1) x^ndx = \left(\sup_{x\in [0,1]} \vert f(x) \vert \right)(1-\delta)^{n+1} \to 0 $$ as $n\to\infty$. For $\varepsilon >0$ there exists $\delta>0$ such that $\sup_{x\in [1-\delta , 1]} \vert f(x) -f(1)\vert \leq \varepsilon$. We have

$$\Big\vert(n+1)\int_{1-\delta}^1 x^n f(x) dx - (1 - (1-\delta)^{n+1})f(1)\Big\vert\\ = \Big\vert(n+1)\int_{1-\delta}^1 x^n f(x) dx - (n+1)\int_{1-\delta}^1 x^nf(1)dx\Big\vert\\ \le(n+1)\int_{1-\delta}^1 x^n \vert f(x) -f(1)\vert dx\le \varepsilon \cdot (n+1)\int_{1-\delta}^1 x^ndx= \varepsilon (1 - (1-\delta)^{n+1})$$ Hence since $(1 - (1-\delta)^{n+1}) \to 1$ $$\limsup_{n\to\infty}\Big\vert(n+1)\int_{0}^1 x^n f(x) dx - f(1)\Big\vert \leq \varepsilon$$ Since this holds for arbitrarily small $\varepsilon>0$ we obtain the desired statement.

Answer 4

Just for fun:

Let $X_n$ be a random variables with densities $p_n (x) = (n+1)x^n$ on $(0,1)$ on a common probability space. Then the goal is to show $$\Bbb E [f(X_n)] \to f(1).$$ The cdf of $X_n$ is $F_n(t) = \int_0^t p_n(x)dx = t^{n+1}$ for $t\in (0,1)$. We have $\Bbb P (\vert 1-X_n\vert > \varepsilon ) = F_n(1-\varepsilon) \to 0$ as $n\to\infty$. This means $X_n \to 1$ in probability. Convergence in probability implies convergence in distribution, therefore $$\Bbb E [f(X_n)] \to f(1).$$