Show $\lim_{n\to\infty}n\int_0^1x^n f(x)dx = 0$, when $f\in L^1([0,1])$, and $\lim_{x\to1^-}f(x)=0$

Question

I thought I had a proof that utilized the dominated convergence theorem, but I realized that I was mistaken in my use of the theorem, and my proof never really used that $f(x)\rightarrow0$ as $x\rightarrow1$.

So all I've really been able to show is that $nx^nf(x)\rightarrow0$ point wise on $[0,1)$. Beyond that I'm stuck.

One thing I've considered is using the fact that $f\in L^1([0,1])$ means that there is some measurable $A\subseteq[0,1]$ such that $m(A)<\epsilon$ and $f$ is bounded on $[0,1]\sim A$ ($\epsilon$ can be arbitrarily small). This would at least give me uniform convergence on $[0,1]\sim A$ and thus prove the claim for $[0,1]\sim A$. But I'm not totally clear on where I would proceed from there, or if it's even the best approach.

Any advice would be greatly appreciated. Thanks in advance.

Answer 1

Let $\varepsilon > 0$. Because $f(x) \rightarrow 0$ as $x \rightarrow 1$, there exists $0 < \eta < 1$ such that for all $x \in [1-\eta, 1]$, $|f(x)| \leq \varepsilon/2$. So you have $$\left| n \int_0^1 x^n f(x) \mathrm{dx}\right| \leq \left|n \int_0^{1-\eta} x^n f(x) \mathrm{dx}\right| + \left|n \int_{1-\eta}^1 x^n f(x) \mathrm{dx}\right|$$

So $$\left| n \int_0^1 x^n f(x) \mathrm{dx}\right| \leq n (1-\eta)^n \int_0^{1-\eta} |f(x)| \mathrm{dx} + n \int_{1-\eta}^1 x^n \frac{\varepsilon}{2} \mathrm{dx}$$

$$ \leq n (1-\eta)^n || f ||_1 + n \frac{\varepsilon}{2} \int_0^1 x^n \mathrm{dx}$$

$$\leq n (1-\eta)^n || f ||_1 + \frac{\varepsilon}{2} \frac{n}{n+1}$$

The first term $n (1-\eta)^n || f ||_1$ tends to $0$, and the second term tends to $\varepsilon /2$, so the whole term tends to $\varepsilon /2$. You deduce that there exists $N \in \mathbb{N}$ such that for $n \geq N$, $$\left| n \int_0^1 x^n f(x) \mathrm{dx}\right| \leq \varepsilon$$

Answer 2

Firstly, we show that for each $c\in[0,1)$, $\lim_{n\rightarrow\infty}\int_{0}^{c}nx^{n}f(x)dx=0$. Observe that \begin{eqnarray*} \left|\int_{0}^{c}nx^{n}f(x)dx\right| & \leq & \int_{0}^{c}nc^{n}|f(x)|dx\\ & \leq & nc^{n}\int_{0}^{1}|f(x)|dx\\ & \rightarrow & 0 \end{eqnarray*} as $n\rightarrow\infty.$ Suppose that $\lim_{x\rightarrow1-}f(x)=\xi$. Let $\varepsilon>0$. Choose $c\in[0,1)$ such that $|f(x)-\xi|<\varepsilon$ whenever $x\in[c,1)$. Choose $N_{1}$ such that $\left|\int_{0}^{c}nx^{n}f(x)dx\right|<\varepsilon$ whenever $n\geq N_{1}$. By direct calculation, we have $\int_{c}^{1}nx^{n}\xi dx=\xi\cdot\frac{n}{n+1}(1-c^{n+1})\rightarrow\xi$ as $n\rightarrow\infty$. Choose $N_{2}$ such that $\left|\int_{c}^{1}nx^{n}\xi dx-\xi\right|<\varepsilon$ whenever $n\geq N_{2}$. Note that \begin{eqnarray*} \int_{c}^{1}|nx^{n}f(x)-nx^{n}\xi|dx & \leq & \int_{c}^{1}nx^{n}\varepsilon dx\\ & = & \varepsilon\cdot\frac{n}{n+1}(1-c^{n+1})\\ & < & \varepsilon. \end{eqnarray*} Now, for any $n\geq\max(N_{1},N_{2})$, we have \begin{eqnarray*} \left|\int_{0}^{1}nx^{n}f(x)dx-\xi\right| & \leq & \left|\int_{0}^{c}nx^{n}f(x)dx\right|+\int_{c}^{1}|nx^{n}f(x)-nx^{n}\xi|dx+\left|\int_{c}^{1}nx^{n}\xi dx-\xi\right|\\ & < & 3\varepsilon. \end{eqnarray*} This show that $\lim_{n\rightarrow\infty}\int_{0}^{1}nx^{n}f(x)dx=\xi:=\lim_{x\rightarrow1-}f(x)$.