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Let $f : [0, 1] → ℝ $ be a continuous function. Show that $$\lim_{n \to \infty} (n + 1) \int_0^1 x^nf(x) \ dx = f(1)$$

Here is an argument...

Let $ε > 0$ be arbitrary. Since $f$ is continuous on $[0, 1]$, it is uniformly continuous on $[0, 1]$. Thus, there exists $δ > 0$ such that $|f(x) - f(y)| < \frac ϵ2 $ for all $x$, $y$ in $[0, 1]$ with $|x - y| < δ$.

Now how to approach...i didn't find any way.. please help..

MathFail
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Abcd
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    I haven't time to check but a trick that often works is to rewrite things as integrals, for example $f(1)=\int_0^1 f(1); dx$. Here, you may use $f(1)=(n+1)\int_0^1 x^nf(1); dx$ and then prove that the limit of $(n+1)\int_0^1x^n(f(x)-f(1)); dx$ is $0$. – Taladris May 07 '23 at 00:57
  • Hint: try to show that the series of general term your expression minus $f(1)$ converges. Once you have convergence of a series, you automatically have the convergence towards $0$ of its general term, and here the latter would give you the desired result. – Bruno B May 07 '23 at 01:04
  • Reminds me very much of this question I answered some time ago. Unfortunately I don't think we can assume differentiability (though I think my argument could be made to work) but Dominated Convergence is probably a good approach. – Rhys Hughes May 07 '23 at 01:22
  • How can I find that this question already has an answer in mathstackexchange...? – Abcd May 07 '23 at 02:24
  • @Abcd: With Approach0. See also How to search with Approach0. – Martin R May 07 '23 at 03:10

1 Answers1

1

$$\forall \epsilon>0, \exists \delta>0, \forall |x-1|<\delta,\Rightarrow f(1)-\epsilon<f(x)<f(1)+\epsilon\tag{1}$$

$$\int_0^1 x^n f(x) dx=\int_0^{1-\delta/2}x^n f(x) dx+\int^1_{1-\delta/2}x^n f(x) dx$$

Since $f$ is continous on $[0, 1-\delta/2]$, hence it attains maxima $M$ and minima $m$.

$$m\int_0^{1-\delta/2}x^n dx\le \int_0^{1-\delta/2}x^n f(x) dx\le M\int_0^{1-\delta/2}x^n dx$$

and

$$\lim_{n\to\infty} ~(n+1)\int_0^{1-\delta/2}x^n dx=\lim_{n\to\infty}\left(1-\frac{\delta}2\right)^{n+1}=0$$

So by squeeze theorem,

$$\lim_{n\to\infty} ~(n+1)\int_0^{1-\delta/2}x^n f(x) dx=0$$

Therefore,

$$\lim_{n\to\infty} ~(n+1)\int_0^1 x^n f(x) dx=\lim_{n\to\infty} ~(n+1)\int^1_{1-\delta/2}x^n f(x) dx=L_2$$

By (1), we have

$$(f(1)-\epsilon)\lim_{n\to\infty} ~(n+1)\int^1_{1-\delta/2}x^n dx\le L_2\le (f(1)+\epsilon)\lim_{n\to\infty} ~(n+1)\int^1_{1-\delta/2}x^n dx$$

Compute the limit,

$$\lim_{n\to\infty} ~(n+1)\int^1_{1-\delta/2}x^n dx=\lim_{n\to\infty} ~1-\left(1-\frac{\delta}2\right)^{n+1}=1$$

Therefore, $$f(1)-\epsilon\le L_2\le f(1)+\epsilon$$

Since $\epsilon $ is arbitrary, we get $$\lim_{n\to\infty} ~(n+1)\int_0^1 x^n f(x) dx=f(1)$$

MathFail
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