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I am told that $ \int_{0}^{1} (n + 1)x^n dx = 1$ and $f$ is continuous on $[0,1]$. I must find

$$\lim_{n \to \infty} \int_{0}^{1} (n + 1)x^n f(x) dx$$

My first impression is that the limit can vary. If $f = 1$, then the limit is just $1$ and if $f = 0$, then the limit is $0$. I was able to bound the whole thing by $\| f \|_\infty$, and so I believe that should be the limit. I have trouble bounding the lower part.

A hint was said to split up the region of $[0,1]$, but i didn't find that useful.

Lemon
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3 Answers3

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Think of $(n+1)x^n$ as a probability density on $[0,1]$. The integral is the average value of $f$ according to that probability density. As $n\to\infty$, the density becomes concentrated near $1$, so we expect that $$ \lim_{n\to\infty} \int_0^1 (n+1)x^n f(x) \,dx = f(1) $$ The idea of splitting up $[0,1]$ is relevant because it allows us to do separate estimates on the part where the density is thin and the part where the density is thick. Something like this: $$ \left|\int_0^{1-\delta} (n+1)x^n f(x) \,dx\right| \le \|f\|_\infty \int_0^{1-\delta} (n+1)x^n \,dx \to 0 \qquad\text{as $n\to\infty$,} $$ and $$ \int_{1-\delta}^1 (n+1) x^n f(x) \,dx \approx f(1) \int_{1-\delta}^1 (n+1) x^n \,dx \to f(1) $$ So choose $\delta$ so that $|f(1)-f(x)|<\varepsilon$ when $1-\delta<x<1$, and try to prove the above.

  • How is $\int_{1-\delta}^1 (n+1) x^n f(x) ,dx \approx f(1) \int_\delta^1 (n+1) x^n ,dx$ justified…? – Lemon Nov 17 '14 at 16:54
  • Sorry, I forgot to fix the $\delta$ in the second integral. I'll edit it. –  Nov 17 '14 at 17:00
  • You'll prove it by using that $f(x)\approx f(1)$ on $[1-\delta,1]$. In general, if $m\le h(x)\le M$ on $[a,b]$ then $m(b-a)\le\int_a^b h(x)\le M(b-a)$. –  Nov 17 '14 at 17:02
  • Sorry i think I got lost in one part. You have shown that your whole integral is $|.| \leq f(1) \int$, how does that show the limit is $f(1)$? That part remains a mystery to me. – Lemon Nov 26 '14 at 05:31
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Since you were asking about Weierstrass Approximation Theorem recently, here's an argument.

Assume first that $f$ is twice continuously differentiable (in particular, a polynomial would satisfy this). Then, integrating by parts, \begin{align} \int_0^1(n+1)x^nf(x)\,dx&=\left.\phantom{\int\!\!\!\!}x^{n+1}f(x)\right|_0^1-\int_0^1x^{n+1}f'(x)\,dx\\ \ \\ &=\left.\phantom{\int\!\!\!\!}x^{n+1}f(x)\right|_0^1-\left.\phantom{\int\!\!\!\!}\frac{x^{n+2}}{n+2}f'(x)\right|_0^1+\int_0^1\frac{x^{n+2}}{n+2}\,f''(x)\,dx\\ \ \\ &=f(1)-\frac{f'(1)}{n+2}+\frac{1}{n+2}\int_0^1 x^{n+2} \,f''(x)\,dx. \end{align} It is easy to see that the last integral is bounded by $\|f''\|_\infty$ (which is finite since we assume $f''$ continuous). So, in the limit, $$ \lim_{n\to\infty}\int_0^1(n+1)x^nf(x)\,dx=f(1). $$

Now, for arbitrary $f$, let $p_n$ be polynomials with $|f-p_n|<1/n$. Then $$ \left|\int_0^1(n+1)x^nf(x)\,dx-p_n(1)\right|=\left|\int_0^1(n+1)x^nf(x)\,dx-\int_0^1(n+1)x^np_n(x)\,dx\right|\\ =\left|\int_0^1(n+1)x^n[f(x)-p_n(x)]\,dx\right|\leq\int_0^1(n+1)x^n|f(x)-p_n(x)|\,dx\\ \leq\frac1n\,\int_0^1(n+1)x^n\,dx=\frac1n. $$ As $p_n(1)\to f(1)$, the above shows that $$ \lim_{n\to\infty}\int_0^1(n+1)x^n\,f(x)\,dx=f(1). $$

Martin Argerami
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Hint: Show that for any continuous function $g:[0,1] \to \Bbb R$, if $g(1) = 0$, then $$ \int_0^1(n+1)x^ng(x)\,dx = 0 $$ from there, it suffices to set $g(x) = f(x) - f(1).$

A nice way to intuit this limit is to note that for each $n$, we're taking a sort of "average". Namely, we're finding $$ \lim_{n \to \infty}\frac{\int_0^1 x^nf(x)\,dx}{\int_0^1 x^n\,dx} $$

Ben Grossmann
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  • Can I ask why are we looking at $x = 1$? – Lemon Nov 17 '14 at 16:26
  • Sure. Consider the function given by $\lim_{n \to \infty} x^n$. Perhaps that will clarify things. – Ben Grossmann Nov 17 '14 at 16:27
  • The limit is $0$ unless $x = 1$. – Lemon Nov 17 '14 at 16:29
  • Do you see why this is significant? – Ben Grossmann Nov 17 '14 at 16:30
  • Perhaps you could explain why $g(1) = 0$ was important? – Lemon Nov 17 '14 at 16:44
  • We could talk about any $f$ directly, but $0$ tends to be a lot easier to work with. Saying $g(1) = 0$ gives you a limit that approaches zero. Or, are you asking in another way why the value at $x = 1$ is generally important here? – Ben Grossmann Nov 17 '14 at 16:50
  • Why $x = 1$ is important here. – Lemon Nov 17 '14 at 16:51
  • The answer is because $(n+1)x^n \to 0$ uniformly on every interval of the form $[0,1 - \epsilon]$ for every $\epsilon > 0$. It follows that $$ \lim_{n \to \infty} \int_0^{1 - \epsilon} (n+1)x^n f(x),dx = 0 $$ no matter what $f$ happens to be. So, $x = 1$ matters because everything else doesn't. – Ben Grossmann Nov 17 '14 at 16:54
  • Well I know that, but I don't understand why on $[1 - \epsilon, 1]$, that the thing tends to $f(1).$ – Lemon Nov 17 '14 at 16:56
  • @jip that's up to you to figure out. Again, I suggest starting with a continuous $g(x)$ satisfying $g(1) = 0$. Show that for every $\epsilon$, we can find a $\delta > 0$ so that $$ \lim_{n \to \infty} \int_{1 - \delta}^1 (n+1)x^n g(x),dx \leq \epsilon $$ Use the continuity of $g$. – Ben Grossmann Nov 17 '14 at 17:01