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Let $f(x)$ , $x \in [0,1]$,be any positive real valued continuous function. Then find $$\lim\limits_{n \rightarrow \infty }(n+1)\int\limits_{0}^{1} x^n f(x)$$

I tried by integrating by parts to get the result $f(1)-(n+1)\int\limits_{0}^{1} x^{n+1}f(x) dx$ But i think this does not help us anyway to find the limit.

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Notice that for any $ 0 < \delta <1$ : $$\lim_{n \rightarrow \infty }(n+1)\int_{0}^{1-\delta} x^n f(x) = 0$$ since it is bounded by $$\ \ (n+1)\sup |f| \cdot (1-\delta)^n.$$

Therefore, for any $\delta >0 $ $$\lim_{n \rightarrow \infty }(n+1)\int_{0}^{1} x^n f(x) = \lim_{n \rightarrow \infty }(n+1)\int_{1-\delta}^{1} x^n f(x) \ .$$

Now for any $n$, $$ \int_{1-\delta}^{1} (n+1) x^n f(x) dx \leq \int_{1-\delta}^{1} (n+1) x^n \sup_{\{1 - \delta \leq y \leq 1 \}} f(y) \ dx \leq [ \sup_{\{1 - \delta \leq y \leq 1 \}} f(y)] \int_{1-\delta}^{1} (n+1) x^n dx $$

Similarly, $$ [ \inf_{\{1 - \delta \leq y \leq 1 \}} f(y)] \int_{1-\delta}^{1} (n+1) x^n dx \leq \int_{1-\delta}^{1} (n+1) x^n f(x) dx \ .$$

Now, letting $n \to \infty$, we get

$$ [ \inf_{\{1 - \delta \leq y \leq 1 \}} f(y)] \leq \int_{1-\delta}^{1} (n+1) x^n f(x) dx \leq [ \sup_{\{1 - \delta \leq y \leq 1 \}} f(y)] \ .$$

This is true for all $\delta$. So, let $\delta \to 0$. Both sides converge to $f(1)$ by contiuity of $f$ at $x=1$. End.

NOTE: This proof shows that continuity at $x=1$ and boundedness of $f$ are sufficient to claim the same argument!

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  • $\begingroup$ This is effectively showing that $\lim_{n\to \infty}(n+1)x^n\sim \delta(x-1)$ in distribution. (+1) $\endgroup$
    – Mark Viola
    Feb 28, 2017 at 18:56
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    $\begingroup$ Huzzah! A rigorous argument using only simple techniques that are available to a student at the point where this exercise is posed. $\endgroup$ Feb 28, 2017 at 19:00
  • $\begingroup$ Can you explain why it is obvious if continuity of $f(x)$ is given? $\endgroup$
    – user321656
    Mar 1, 2017 at 2:29
  • $\begingroup$ Because by definition of continuity at $x=1$ the quantity $\sup_{\{0<1-x<\delta\}} |f(x)-f(1)|$ goes to zero if $\delta$ goes to zero. The $\delta-\epsilon$ definition of continuity. $\endgroup$ Mar 1, 2017 at 6:27
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Let us assume that $f \in C^1([0,1])$. We then have $$(n+1)\int_0^1 x^n f(x)dx = \int_0^1 f(x) d\left(x^{n+1}\right) = f(1) - \int_0^1x^{n+1}f'(x)dx = f(1) - f'(y)\int_0^1 x^{n+1}dx$$ for some $y \in [0,1]$. This so since $f'$ is continuous over $[0,1]$. Hence, we obtain that $$\lim_{n \to \infty} (n+1)\int_0^1 x^n f(x)dx = f(1)$$


From the above, we see that the result is true if $f(x)$ happens to be a polynomial.

Given any continuous function $f(x)$ from Weierstrass approximation theorem, $f(x)$ can be uniformly approximated by polynomials on a bounded interval.

This can be easily proved using Bernstein polynomials, uniform continuity, boundedness of continuous function on compact sets and Chebyshev inequality.

Hence, given any $\epsilon > 0$, there exists a polynomial $p_{\epsilon}(x)$ such that $$\left \vert f(x)-p_{\epsilon}(x) \right \vert < \epsilon \,\,\,\,\,\, \forall x \in [0,1]$$

Hence, we have that $$\left \vert (n+1)\int_0^1 x^n\left(f(x)-p_{\epsilon}(x) \right)dx\right \vert \leq \epsilon (n+1)\int_0^1 x^n dx = \epsilon$$ for all $n$. Hence, for any $\epsilon>0$, we obtain that $$\lim_{n \to \infty} (n+1)f(x)dx \in \left(p_{\epsilon}(1)-\epsilon,p_{\epsilon}(1)+\epsilon\right)$$ Since this is true for all $\epsilon$ and since $\lim_{\epsilon \to 0}p_{\epsilon}(x) = f(x)$, we obtain that $$\lim_{n \to \infty} (n+1)f(x)dx = f(1)$$

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A more low-tech approach than convergence in distribution$^{(*)}$.
Since $f(x)$ is a continuous function on $[0,1]$, it can be uniformly approximated by polynomials by Weierstrass approximation theorem. It follows that it is enough to prove the claim $$ \lim_{n\to +\infty}\int_{0}^{1} (n+1)\,x^{n} f(x)\,dx = f(1) $$ assuming that $f$ is a polynomial. On the other hand, if $$ f(x) = \sum_{k=0}^{M} a_k x^k $$ we have: $$ \lim_{n\to +\infty}\int_{0}^{1}(n+1)x^n\,f(x)\,dx = \lim_{n\to +\infty}\sum_{k=0}^{M}\frac{(n+1) a_k}{(k+n+1)}=\sum_{k=0}^{M}a_k=f(1) $$ and we are done.


$(*)$ $ g_n(x) = (n+1)x^n $ is a continuous function on $[0,1]$ with unit integral. Since $\{g_n(x)\}_{n\geq 1}$ is pointwise convergent to zero on $(0,1)$, $g_n(x)$ converges to $\delta(x-1)$ in distribution, and $$ \lim_{n\to +\infty}\int_{0}^{1} g_n(x)\,f(x)\,dx = \int_{0}^{1} \delta(x-1)\,f(x)\,dx = f(1).$$

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    $\begingroup$ Yes, nice application of WAT. (+1) (although I liked your appealing to the regularized Dirac Delta in your first post). Approximation of $f$ in the $L^1$ norm by a smooth function is one way of proving the Riemann Lebesgue Lemma too(we also need there that the approximating function is of compact support). $\endgroup$
    – Mark Viola
    Feb 28, 2017 at 18:28
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Depending on your background, one way to approach the question is to assume first that $f$ is polynomial and then approximate $f$ by a polynomial and use a standard $3\varepsilon$ argument. Namely, if $f(x) = a_0 + \dots + a_m x^m$ then we can calculate

$$(n+1) \int_0^1 x^n f(x) \, dx = (n+1) \int_0^1 a_0 x^n + a_1 x^{n+1} + \dots + a_m x^{n + m} \, dx = \\ (n+1) \left[ a_0 \frac{x^{n + 1}}{n+1}+ \dots + a_m \frac{x^{n + m + 1}}{n + m + 1} \right]_{x = 0}^{x = 1} = a_0 + a_1 \frac{n+1}{n+2} + \dots + a_m \frac{n + 1}{n + m + 1} $$

which tends to $f(1) = a_0 + a_1 + \dots + a_m$ as $n \to \infty$.

Now, Weierstrass tells us that given $\varepsilon > 0$, we can approach $f$ uniformly by a polynomial $p$ with $\sup_{x \in [0,1]} |f(x) - p(x)| < \varepsilon$. Then

$$ \left| (n+1) \int_0^1 x^n f(x) \, dx - f(1) \right| = \\ \left| (n+1) \int_0^1 x^n (f(x) - p(x)) \, dx + (n+1) \int_0^1 x^n p(x) \, dx - p(1) + p(1) - f(1) \right| \leq \\ (n+1) \int_0^1 x^n \varepsilon \, dx + \left| (n+1) \int_0^1 x^n p(x) \, dx - p(1) \right| + |p(1) - f(1)| \leq \\ 2\varepsilon + \left| (n+1) \int_0^1 x^n p(x) \, dx - p(1) \right|. $$

The second term will be $\leq \varepsilon$ for $n$ large enough (we showed this by showing the result for polynomials) and so we see that for $n$ large enough (depending on $\varepsilon$) we can make the difference be less than $3\varepsilon$, showing the result for all continuous functions (positive or not).

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  • $\begingroup$ Even if you can't use approximation by polynomials, the approach of assuming first that $p$ is a polynomial suggests to you what should be the "correct limit" and so gives you something to work on. $\endgroup$
    – levap
    Feb 28, 2017 at 18:15

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