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I am trying to solve this problem from a past exam.

Let $f:[0,1]\rightarrow\mathbb R$ be a continuous function such that $f(0)=0$ and $f(1)=1$. Evaluate the limit $$ \lim_{n\rightarrow\infty}n\int_0^1f(x)x^{2n}dx. $$

Since $nf(x)x^{2n}$ is not uniformly convergent on [0,1] (even on [0,1)), I cannot swap $\lim$ and $\int$. I could use integration by parts, but I would get a horrendous alternating sum that involves repeated antiderivatives.

I would appreciate if you could give a clue to this problem.

EDIT: While Peter Tamaroff's answer works and is elegant, I am also looking forward for a solution that one can easily come up with.

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$K_{n}(x) = (2n+1)x^{2n}$ is an approximation to the identity on $[0, 1]$ as $n \to \infty$. That is, $K_{n}(x) \to \delta(x-1)$ at least some sense, and checking this is not hard as Peter Tamaroff did. – Sangchul Lee Aug 14 '13 at 4:04
The hypothesis that $f(0)=0$ seems irrelephant. – Pedro Tamaroff Aug 14 '13 at 4:08
@sos440 Could you please elaborate on that? – Pteromys Aug 14 '13 at 6:08
up vote 11 down vote accepted

I'd rather evaluate $$\left( {2n + 1} \right)\int_0^1 f (x){x^{2n}}dx$$ since $$\int_0^1 {{x^{2n}}dx} = \frac{1}{{2n + 1}}$$

Note then that $$(2n+1)\int_0^1 {f\left( x \right){x^{2n}}dx} - f\left( 1 \right) = \left( {2n + 1} \right)\int_0^1 ({f\left( x \right) - f\left( 1 \right)){x^{2n}}dx} $$

and that $(2n+1)x^{2n}$ converges uniformly to zero on $[0,\delta)$ for any $\delta >0$. Use that $f(x)-f(1)$ is continuous and is zero at $x=1$ to make the integral small in a neighborhood $[\delta,1]$. If the above isn't clear enough:

$$\begin{align}\left|\int_0^1(f(x)-f(1))x^{2n}dx\right|&\leqslant\int_0^1|f(x)-f(1)|x^{2n}dx\\&=\int_0^\delta|f(x)-f(1)|x^{2n}dx+\int_\delta^1|f(x)-f(1)|x^{2n}dx\\&\leqslant M\int_0^\delta x^{2n}dx+\varepsilon\int_\delta^1x^{2n}\\&\leqslant M\int_0^\delta x^{2n}dx+\varepsilon\cdot \int_0^1x^{2n}\\&=M \int_0^\delta x^{2n}dx+\frac{\epsilon}{2n+1}\end{align}$$

Upon multiplication by $2n+1$, you get the limit in question must indeed be $f(1)$. Thus your limit is $f(1)/2$.

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grrrrr beat me to it – Euler....IS_ALIVE Aug 14 '13 at 3:52
@Peter Do you mean in the third equation $(2n+1)\int \dots dx - f(1) = \dots$? – Pteromys Aug 14 '13 at 5:57
@peter-tamaroff Could you please explain why $(2n+1)x^{2n}$ is uniformly convergent? I might be mistaken, but if one considers the quantity $\Delta_{n,m}(x):=|(2n+1)x^{2n}−(2m+1)x^{2m}|$ to use "Cauchy's criterion", one has $\Delta_{n,m}(x)=x^m|(2n+1)x^{2(n−m)}−(2m+1)|$ and this do not tend to zero as $n\rightarrow\infty$ where $m$ fixed, no matter how large $m$ is, which means the function sequence does not converge uniformly. – Pteromys Aug 14 '13 at 10:46
@Chris'ssis Well, I am saying I am evaluating $$\left( {2n + 1} \right)\int_0^1 f (x){x^{2n}}dx$$ so if this is $f(1)$ then you can see $n\int_0^1 f (x){x^{2n}}dx$ is $f(1)/2$. You should read things more carefully next time. – Pedro Tamaroff Aug 15 '13 at 11:22
@Chris'ssis You don't need to downvote anything. Personally, my philosophy (in the past one year, at least) has been never to downvote but just to upvote as many good faith answers as I can. If there's an answer that you think is deficient, then why not post a comment. Downvotes have a purpose, I suppose, when the user is clearly contributing in bad faith but that's clearly not the case here (Peter is an excellent contributor). I think downvoting good faith, knowledgeable and earnest contributors in this way is negative and doesn't contribute to the development of this site. – Amitesh Datta Aug 16 '13 at 2:08

\begin{eqnarray*} \lim_{n \to \infty}n\int_{0}^{1}{\rm f}\left(x\right)x^{2n}\,{\rm d}x & = & \lim_{n \to \infty}\left\lbrack% n\int_{0}^{1}{\rm f}\left(1 - \epsilon\right)\left(1 - \epsilon\right)^{2n}\,{\rm d}\epsilon \right\rbrack \\[3mm] & = & \lim_{n \to \infty}\left\lbrack n\int_{0}^{1} {\rm e}^{\ln\left(\vphantom{\LARGE A}{\rm f}\left(1\ -\ \epsilon\right)\right)\ +\ 2n\ln\left(1\ -\ \epsilon\right)}\,{\rm d}\epsilon\right\rbrack \\[3mm] & = & \lim_{n \to \infty}\left\lbrace n\int_{0}^{1} {\rm e}^{\ln\left(\vphantom{\LARGE A}{\rm f}\left(1\right)\right)\ -\ \left\lbrack\vphantom{\LARGE A}{\rm f}'\left(1\right)/{\rm f}\left(1\right)\right\rbrack\epsilon\ -\ 2n\epsilon}\,{\rm d}\epsilon\right\rbrace \\[3mm] & = & {\rm f}\left(1\right)\lim_{n \to \infty}\left\lbrack n\ \times\ {{\rm e}^{-{\rm f}'\left(1\right)/{\rm f}\left(1\right)\ -\ 2n} - 1 \over -{\rm f}'\left(1\right)/{\rm f}\left(1\right) -2n}\right\rbrack = {\rm f}\left(1\right)\,{1 \over 2} = {1 \over 2} \end{eqnarray*}

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