Let $f$ be continuous on $[0,1]$. Find the limit $\lim_{n \rightarrow \infty}(n+1) \sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} f(x) \mathrm{d} x$.

Question

$\int_0^1 x^k (1-x)^{n-k} dx$ is a Beta function.

The probability density function of the Beta distribution is given by $f(x; \alpha, \beta) = \frac{1}{B(\alpha, \beta)} x^{\alpha - 1} (1 - x)^{\beta - 1}$.

Therefore, $x^k (1-x)^{n-k} = B(\alpha, \beta)\frac{1}{B(\alpha, \beta)} x^{\alpha - 1} (1 - x)^{\beta - 1}$ can be seen as the density function of a Beta distribution multiplied by $B(\alpha, \beta)$, where $\alpha = k + 1$ and $\beta = n - k + 1$.

Hence, we can rewrite the original limit as $$ \lim_{n \rightarrow \infty} (n+1) \sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} f(x) \mathrm{d} x \sim\lim_{n \rightarrow \infty} (n+1) \sum_{k=0}^n \text{Beta}(k+1, n-k+1)\mathbb{E}_{X \sim \text{Beta}(k+1, n-k+1)}[f(X)], $$ But I don't know how to deal with it next.t_t

Answer 1

We have $$ \sum_{k=0}^nx^k(1-x)^{n-k}=(1-x)^n{1-{x^{n+1}\over (1-x)^{n+1}}\over 1-{x\over 1-x}}\\ = {(1-x)^{n+1}-x^{n+1}\over 1-2x}$$ For $f(x)=(1-2x)g(x)$, where $g(x)$ is continuous for $x\neq 1/2$ and absolutely integrable, we obtain $$(n+1)\int\limits_0^1(1-x)^{n+1}g(x)\,dx\\ - (n+1)\int\limits_0^1x^{n+1}g(x)\,dx $$ The limit is equal $$g(0)-g(1) =f(0)+f(1)$$ as in general $$\lim_n(n+1)\int\limits_0^1x^nh(x)\,dx =h(1)$$ The latter follows easily from the fact that $(n+1)x^n$ tends to $0$ uniformly on each interval $[0,1-\delta]$ and $(n+1)\int\limits_0^1x^n\,dx=1.$

In general if $f(x)/|2x-1|$ is not absolutely integrable we can approximate $f(x)$ by $f_k(x)=f(x)|2x-1|^{1/k}$ for which the ratio $f_k(x)/|2x-1|$ is absolutely integrable. As $f_k(0)=f(0)$ and $f_k(1)=f(1)$ the limit is equal $f(0)+f(1).$

Answer 2

The answer is $$\lim_{n \rightarrow \infty}(n+1) \sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} f(x) \,\mathrm{d} x=f(0)+f(1).$$

Denote $$I_n= n\sum_{k=0}^{n-1}\int_0^1x^k(1-x)^{n-1-k}f(x)\,\mathrm{d} x,\qquad n\in\mathbb N_{\geq1}.$$ Using the summation formula for geometric series we obtain $$I_n=n\int_0^1\frac{(1-x)^n-x^n}{1-2x}f(x)\,\mathrm{d} x.$$ Let $x=(1+s)/2$, then $$I_n=\frac n{2^{n+1}}\int_{-1}^1\frac{(1+s)^n-(1-s)^n}{s}g(s)\,\mathrm{d} s,$$ where $g(s)=f((1+s)/2)$ is a continuous function on $[-1,1]$, so $M:=\sup_{s\in[-1,1]}|g(s)|<+\infty$.

For each $\varepsilon\in(0,1/2)$, there exists $\delta\in(0, \varepsilon)$ such that $|g(s)-g(1)|<\varepsilon$ for all $s\in(1-\delta, 1)$ and $|g(s)-g(-1)|<\varepsilon$ for all $s\in(-1, -1+\delta)$. Now we decompose $I_n$ into several parts: let \begin{align*} I_n^{(1)}&=\frac n{2^{n+1}}\int_{-1+\delta}^{1-\delta}\frac{(1+s)^n-(1-s)^n}{s}g(s)\,\mathrm{d} s\\ I_n^{(2)}&=\frac n{2^{n+1}}\int_{-1}^{-1+\delta}\frac{(1+s)^n-(1-s)^n}{s}(g(s)-g(-1))\,\mathrm{d} s\\ I_n^{(3)}&=\frac n{2^{n+1}}\int_{1-\delta}^1\frac{(1+s)^n-(1-s)^n}{s}(g(s)-g(1))\,\mathrm{d} s\\ I_n^{(4)}&=g(-1)\frac n{2^{n+1}}\int_{-1}^{-1+\delta}\frac{(1+s)^n-(1-s)^n}{s}\,\mathrm{d} s\\ I_n^{(5)}&=g(1)\frac n{2^{n+1}}\int_{1-\delta}^1\frac{(1+s)^n-(1-s)^n}{s}\,\mathrm{d} s, \end{align*} then $I_n=I_n^{(1)}+I_n^{(2)}+I_n^{(3)}+I_n^{(4)}+I_n^{(5)}$. For $I_n^{(1)}$, by the mean value theorem, for any $s\in(-1+\delta, 1-\delta)$, there exists $\xi\in(-1+\delta, 1-\delta)$ such that $(1+s)^n-(1-s)^n=2n(1+\xi)^{n-1}s$, and thus $$\left|(1+s)^n-(1-s)^n\right|\leq 2ns(2-\delta)^{n-1}, \qquad s\in(-1+\delta, 1-\delta).$$ Hence $$\left|I_n^{(1)}\right|\leq \frac n{2^{n+1}}\cdot 2n(2-\delta)^{n-1}M\cdot 2(1-\delta)\leq Mn^2\left(\frac{2-\delta}2\right)^n\rightarrow0, \text{as }n\rightarrow\infty, $$ and thus for $n$ sufficiently large, we have $\left|I_n^{(1)}\right|\leq \varepsilon$. For $I_n^{(j)}(2\leq j\leq 5)$, we just need to estimate $$\frac n{2^{n+1}}\int_{1-\delta}^1\frac{(1+s)^n-(1-s)^n}{s}\,\mathrm{d} s.$$ Clearly we have $$\int_{1-\delta}^1\left[(1+s)^n-(1-s)^n\right]\,\mathrm{d} s\leq\int_{1-\delta}^1\frac{(1+s)^n-(1-s)^n}{s}\,\mathrm{d} s\leq \frac1{1-\delta}\int_{1-\delta}^1(1+s)^n\,\mathrm{d} s,$$ hence $$\frac{2^{n+1}-(2-\delta)^{n+1}-\delta^{n+1}}{n+1}\leq\int_{1-\delta}^1\frac{(1+s)^n-(1-s)^n}{s}\,\mathrm{d} s\leq\frac1{1-\delta} \frac{2^{n+1}-(2-\delta)^{n+1}}{n+1},$$ and then $$1\leq\liminf_{n\to\infty}\frac n{2^{n+1}}\int_{1-\delta}^1\frac{(1+s)^n-(1-s)^n}{s}\,\mathrm{d} s\leq\limsup_{n\to\infty}\frac n{2^{n+1}}\int_{1-\delta}^1\frac{(1+s)^n-(1-s)^n}{s}\,\mathrm{d} s\leq \frac1{1-\delta}.$$ So, for $n$ sufficiently large, we have $$\left|I_n^{(2)}\right|+\left|I_n^{(3)}\right|\leq \frac{4\varepsilon}{1-\delta}\leq 8\varepsilon,$$ and $$\left|I_n^{(4)}-g(-1)\right|+\left|I_n^{(5)}-g(1)\right|\leq 4M\left(\frac1{1-\delta}-1\right)=\frac{4M\delta}{1-\delta}\leq 8M\varepsilon.$$

Now we have shown that for each $\varepsilon\in(0,1/2)$, there holds $$\left|I_n-(g(-1)+g(1))\right|\leq \varepsilon+8\varepsilon+8M\varepsilon=(8M+9)\varepsilon,$$ for $n$ sufficiently large. Therefore $$\lim_{n \rightarrow \infty} I_n=g(-1)+g(1)=f(0)+f(1).$$

Answer 3

First, we prove that for any polynomial $P(x)$, the limit $$ \lim_{n \to \infty} (n+1) \sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} P(x) \mathrm{d} x =P(0)+P(1) $$ is correct.

Without loss of generality，let $P(x)=x^m\ (m=0,1,2,\cdots)$, and we use the $\beta $ function, then we have $$ \begin{aligned} \lim_{n\to\infty}(n+1)\sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} P(x) \mathrm{d} x &=\lim_{n\to\infty}(n+1)\sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} x^m \mathrm{d} x \\ &=\lim_{n\to\infty}(n+1)\sum_{k=0}^n \frac{1}{k+m}\frac{1}{\displaystyle{\binom{n+m}{k+m}}}\\ &=1 \end{aligned} $$ The final limit is based on the characteristic of combination, which is $$ \binom{b}{a} \geq \frac{b(b-1)}{2}\ ,\ 2 \leq a \leq b-2 $$ We note that the polynomial space is a linear space , so for any polynomial $P(x)$, the limit $$ \lim_{n \to \infty} (n+1) \sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} P(x) \mathrm{d} x =P(0)+P(1) $$ is established.

Secondly, we use the $weierstrass-stone$ theory, which is, for the continuous function f on the closed interval $[0,1]$, and for any fixed $\varepsilon > 0$, then there exsits a polynomial $P_0(x)$, satisfied $$ \sup_{x \in [0,1]} |f(x)-P_0(x)| < \varepsilon $$ Then $$ \begin{aligned} (n+1) \sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} f(x) \mathrm{d} x &\leq (n+1) \sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} (P_0(x) + \varepsilon) \mathrm{d} x\\ &=\sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} P_0(x) \mathrm{d} x + \varepsilon (n+1) \sum_{k=0}^n \frac{1}{\displaystyle{\binom{n}{k}}} \end{aligned} $$ Use the characteristic of combination, then $$ \lim_{n\to\infty} \sum_{k=0}^n \frac{1}{\displaystyle{\binom{n}{k}}}=2 $$ So we have $$ \begin{aligned} \varlimsup_{n\to\infty}(n+1) \sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} f(x) \mathrm{d} x &\leq P_0(0) + P_0(1) + 2 \varepsilon\\ &\leq f(0) + f(1) + 4 \varepsilon \end{aligned} $$ Use the same method, we have $$ \varliminf_{n\to\infty}(n+1) \sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} f(x) \mathrm{d} x \geq f(0) + f(1) +4\varepsilon $$ Finally, let $\varepsilon \to 0$, the we have $$ \lim_{n\to\infty}(n+1) \sum_{k=0}^n \int_0^1 x^k(1-x)^{n-k} f(x) \mathrm{d} x =f(0)+f(1) $$

Answer 4

Let $\,f:[0,1]\to\Bbb R\,$ be a continuous function on $\,[0,1]\,.$ Find the limit $\,\displaystyle\lim\limits_{n\to\infty}\!\left[\big(n+1\big)\!\sum_{k=0}^n\int_0^1x^k(1-x)^{n-k}f(x)\,\mathrm dx\right]\!.$

Since $\,f\,$ is continuous on $\,[0,1]\,,\,$ there exists $\,M>0\,$ such that $\,\left|f(x)\right|\leqslant M\;\;$ for any $\,x\in[0,1]\,.\quad\color{blue}{(1)}$

Moreover, for any $\;n\in\Bbb N\,\land\,n\geqslant3\,,\,$ it results that

$\displaystyle\left|\big(n+1\big)\!\sum_{k=1}^{n-1}\int_0^1x^k(1-x)^{n-k}f(x)\,\mathrm dx\right|\leqslant$

$\displaystyle\leqslant\big(n+1\big)\!\sum_{k=1}^{n-1}\left|\int_0^1x^k(1-x)^{n-k}f(x)\,\mathrm dx\right|\leqslant$

$\displaystyle\leqslant\big(n+1\big)\!\sum_{k=1}^{n-1}\int_0^1x^k(1-x)^{n-k}\big|f(x)\big|\,\mathrm dx\underset{\overbrace{\;\text{ by applying }(1)\;}}{\leqslant}$

$\displaystyle\leqslant\big(n+1\big)M\!\sum_{k=1}^{n-1}\int_0^1x^k(1-x)^{n-k}\,\mathrm dx=$

$\displaystyle=\big(n+1\big)M\!\sum_{k=1}^{n-1}\mathrm{Beta}\big(k+1,n-k+1\big)=$

$\displaystyle=\big(n+1\big)M\!\sum_{k=1}^{n-1}\frac{k!\big(n-k\big)!}{\big(n+1\big)!}=M\!\sum_{k=1}^{n-1}\frac1{\binom nk}\underset{\overbrace{\,\text{because }\binom nk\geqslant\binom n2\\\,\text{for any}\,2\leqslant k\leqslant n-2\,}}{\leqslant}$

$\leqslant\!M\!\!\left[\dfrac1{\binom n1}+\dfrac{n-3}{\binom n2}+\dfrac1{\binom n{n-1}}\right]\!=M\!\left[\dfrac2n+\dfrac{2(n-3)}{n(n-1)}\right]\!<\dfrac{4M}n\,.$

Consequently, by applying Squeeze theorem, it follows that

$\displaystyle\lim\limits_{n\to\infty}\!\left[\big(n+1\big)\!\sum_{k=1}^{n-1}\int_0^1x^k(1-x)^{n-k}f(x)\,\mathrm dx\right]\!=0\;.\quad\color{blue}{(2)}$

Now, we will prove that

$\displaystyle\lim\limits_{n\to\infty}\!\left[\big(n+1\big)\!\int_0^1x^nf(x)\,\mathrm dx\right]\!=f(1)\;.$

For any $\,\varepsilon>0\,,\,$ since $\,\lim\limits_{x\to1}f(x)=f(1)\,,\,$ there exists $\,x_\varepsilon\in(0,1)\,$ such that $\;\big|f(x)-f(1)\big|<\dfrac\varepsilon2\;\;\;\forall x\in\left[x_\varepsilon,1\right].\quad\color{blue}{(3)}$

Moreover,
$\big|f(x)-f(1)\big|\leqslant\big|f(x)\big|+\big|f(1)\big|\leqslant2M\;\;\;\forall x\in\left[0,x_\varepsilon\right].\quad\color{blue}{(4)}$

Let $\;n_\varepsilon=\max\left\{1,\left\lfloor\dfrac{\ln\varepsilon-\ln(4M)}{\ln x_\varepsilon}\right\rfloor\!-\!1\right\}\in\Bbb N\,.$

For any $\,n\in\Bbb N\,\land\,n>n_\varepsilon\,$ it results that

$2Mx_\varepsilon^{n+1}\leqslant2Mx_\varepsilon^{n_\varepsilon+2}\leqslant2Mx_\varepsilon^{\left\lfloor\frac{\ln\varepsilon-\ln(4M)}{\ln x_\varepsilon}\right\rfloor+1}<2Mx_\varepsilon^{\frac{\ln\varepsilon-\ln(4M)}{\ln x_\varepsilon}}=$
$=2Mx_\varepsilon^{\log_{x_\varepsilon}\left(\frac{\varepsilon}{4M}\right)}=2M\dfrac{\varepsilon}{4M}=\dfrac{\varepsilon}2\;\;,\quad\color{blue}{(5)}$

$\displaystyle\left|\big(n+1\big)\!\int_0^1x^nf(x)\,\mathrm dx-f(1)\right|=$

$\displaystyle=\left|\int_0^1(n+1)x^nf(x)\,\mathrm dx-\int_0^1(n+1)x^nf(1)\mathrm\,dx\right|=$

$\displaystyle=\left|\int_0^1(n+1)x^n\big[f(x)-f(1)\big]\,\mathrm dx\right|\leqslant$

$\displaystyle\leqslant\int_0^1(n+1)x^n\big|f(x)-f(1)\big|\,\mathrm dx=$

$\displaystyle=\!\int_0^{x_\varepsilon}\!\!\!(n\!+\!1)x^n\big|f(x)\!-\!f(1)\big|\,\mathrm dx+\!\!\int_{x_\varepsilon}^1\!(n\!+\!1)x^n\big|f(x)\!-\!f(1)\big|\,\mathrm dx<$

$\displaystyle\underset{\overbrace{(3)\text{ and }(4)}}{<}2M\!\!\int_0^{x_\varepsilon}\!\!\!(n\!+\!1)x^n\mathrm dx+\dfrac{\varepsilon}2\!\int_{x_\varepsilon}^1\!(n\!+\!1)x^n\mathrm dx=$

$=2M\big[x^{n+1}\big]_0^{x_\varepsilon}+\dfrac{\varepsilon}2\big[x^{n+1}\big]_{x_\varepsilon}^1=$

$=2Mx_{\varepsilon}^{n+1}+\dfrac{\varepsilon}2\left(1-x_{\varepsilon}^{n+1}\right)\underset{\overbrace{\text{ by applying }(5)\;}}{<}\dfrac\varepsilon2+\dfrac\varepsilon2=\varepsilon\,.$

So we have proved that for any $\,\varepsilon>0\,,\,$ there exists $\,n_{\varepsilon}\in\Bbb N\,$ such that for any $\,n\in\Bbb N\,\land\,n>n_\varepsilon\,$ it results that
$\displaystyle\left|\big(n+1\big)\!\int_0^1x^nf(x)\,\mathrm dx-f(1)\right|<\varepsilon\;\,,\;\;\;\;$ that is ,

$\displaystyle\lim\limits_{n\to\infty}\!\left[\big(n+1\big)\!\int_0^1x^nf(x)\,\mathrm dx\right]\!=f(1)\;.\quad\color{blue}{(6)}$

Analogously, we can prove that

$\displaystyle\lim\limits_{n\to\infty}\!\left[\big(n+1\big)\!\int_0^1x^ng(x)\,\mathrm dx\right]\!=g(1)\qquad\color{blue}{(7)}$

for any continuous function $\,g\,$ on $\,[0,1]\,.$

In particular, if $\,g:[0,1]\to\Bbb R\,$ is the continuous function defined as $\,g(t)=f(1-t)\;$ for any $\,t\in[0,1]\,,\,$ by applying the change of variable $\,t=1-x\,$ in the following definite integral, we get that

$\displaystyle\lim\limits_{n\to\infty}\!\left[\big(n+1\big)\!\int_0^1(1-x)^nf(x)\,\mathrm dx\right]=$

$\displaystyle=\lim\limits_{n\to\infty}\!\left[\big(n+1\big)\!\int_0^1t^nf(1-t)\,\mathrm dt\right]\!=$

$\displaystyle=\lim\limits_{n\to\infty}\!\left[\big(n+1\big)\!\int_0^1t^ng(t)\,\mathrm dt\right]\!=g(1)=f(1-1)=f(0)\;.$

Consequently ,

$\displaystyle\lim\limits_{n\to\infty}\!\left[\big(n+1\big)\!\int_0^1(1-x)^nf(x)\,\mathrm dx\right]=f(0)\;.\quad\color{blue}{(8)}$

From $\,(8)\,,\,$ $(2)\,$ and $\,(6)\,,\,$ it follows that

$\displaystyle\lim\limits_{n\to\infty}\!\left[\big(n+1\big)\!\sum_{k=0}^n\int_0^1x^k(1-x)^{n-k}f(x)\,\mathrm dx\right]=f(0)+f(1)\,.$

Answer 5

Let $f : [0, 1] \to \mathbb{R}$ be continuous with the supremum norm $\|f\| := \sup_{[0,1]} |f|$.

Observation 1. The combined contribution from the terms for $0 < k < n$ is negligible. Indeed, it is well-known that

$$ I^{(n)}_{k} := (n+1) \int_{0}^{1} x^k (1-x)^{n-k} \, \mathrm{d}x = \frac{1}{\binom{n}{k}}. $$

From this and the basic properties of the binomial coefficient, we observe that

\begin{gather*} I^{(n)}_{1} = I^{(n)}_{n-1} = \frac{1}{n} \qquad\text{and}\qquad I^{(n)}_{k} \leq \frac{1}{\binom{n}{2}} \quad \text{for $2 \leq k \leq n-2$}. \end{gather*}

Plugging this to OP's sum restricted to the range $0 < k < n$, we obtain

\begin{align*} &\left| (n+1) \sum_{k : 0<k<n} \int_{0}^{1} x^k (1-x)^{n-k} f(x) \, \mathrm{d}x \right| \\ &\qquad \leq \sum_{k : 0<k<n} \|f\| I^{(n)}_{k} \leq \|f\| \biggl( \frac{2}{n} + \frac{n-3}{\binom{n}{2}} \biggr) \to 0. \end{align*}

Observation 2. When $k = n$, the substitution $y = x^{n+1}$ gives

\begin{align*} \int_{0}^{1} (n+1) x^{n} f(x) \, \mathrm{d}x &= \int_{0}^{1} f(y^{\frac{1}{n+1}}) \, \mathrm{d}y \to \int_{0}^{1} f(1) \, \mathrm{d}y = f(1), \end{align*}

where we can utilize the dominated convergence theorem (or the standard approximation-to-the-identity argument) to justify the limit step. By a similar argument, when $k = 0$, we get

\begin{align*} \int_{0}^{1} (n+1) (1-x)^{n} f(x) \, \mathrm{d}x \to f(0). \end{align*}

Conclusion. Combining altogether, we conclude:

$$ \bbox[border:1px dotted navy;padding:5px;color:navy;]{\lim_{n\to\infty} (n+1) \sum_{k=0}^{n} \int_{0}^{1} x^k (1-x)^{n-k} f(x) \, \mathrm{d}x = f(0) + f(1)} $$

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Addendum. As per @Angelo's request, I included an elementary argument implementing approximation-to-the-identity:

Fix $\varepsilon > 0$. By the continuity of $f$ at $x = 1$, there exists $\delta > 0$ such that $|f(x) - f(1)| \leq \varepsilon$ whenever $|x - 1| \leq \delta$. Then

\begin{align*} &\left| \int_{0}^{1} (n+1) x^{n} f(x) \, \mathrm{d}x - f(1) \right| \\ &= \left| \int_{0}^{1} (n+1) x^{n} [f(x) - f(1)] \, \mathrm{d}x \right| \\ &\leq \varepsilon \left( \int_{1-\delta}^{1} (n+1) x^{n} \, \mathrm{d}x \right) + 2\|f\| \left( \int_{0}^{1-\delta} (n+1) x^{n} \, \mathrm{d}x \right) \\ &\leq \varepsilon + 2\|f\| (1-\delta)^{n+1}. \end{align*}

Taking $\limsup$ as $n\to\infty$, we get

$$ \limsup_{n\to\infty} \left| \int_{0}^{1} (n+1) x^{n} f(x) \, \mathrm{d}x - f(1) \right| \leq \varepsilon. $$

Since the limsup is independent of the choice of $\varepsilon$, letting $\varepsilon \to 0^+$ shows that the limsup is zero. Therefore we conclude

$$ \int_{0}^{1} (n+1) x^{n} f(x) \, \mathrm{d}x \to f(1). $$