Let $f \in L^{1}([0,1])$ be a real valued function. Prove the following

Question

Let $f \in L^{1}([0,1])$ be a real valued function. Prove the following

$1) x^k f(x) \in L^1([0,1])$ for all $k\in \mathbb{N}$

$2) \lim_{k\rightarrow\infty}\int_{0}^{1}x^k f(x) dx = 0$

$3)$ If $\lim_{x\rightarrow1}f(x) = a$ for some real number a, then

$\lim_{k\rightarrow\infty} k \int_{0}^{1}x^k f(x) dx = a $

I am not able to do part $3)$

I did $1)$ by comparing with $f$ and $2)$ by dominated convergence theorem. I would really appreciate a hint for $3)$

Answer 1

Expanding on Did's comment, let $\epsilon > 0$ and let $\delta > 0$ be such that if $1 - \delta < x < 1$ then $|f(x) - a| < \epsilon$.

Notice that $\int_0^1kx^k\,dx = 1$. Then \begin{align} \Big|\int_0^1kx^kf(x)\,dx - a\Big| = &\ \Big|\int_0^1kx^k(f(x) - a)\,dx\Big|\\ \le &\ \int_0^{1 - \delta}kx^k|f(x) - a|\,dx + \int_{1 - \delta}^1kx^k|f(x) - a|\,dx \\ \le &\ \int_0^{1 - \delta}kx^k|f(x) - a|\,dx + \epsilon\int_{1 - \delta}^1kx^k \\ \le &\ \int_0^{1 - \delta}kx^k|f(x) - a|\,dx + \epsilon \\ \to &\ 0 + \epsilon. \end{align}

This shows that $\lim_{k \to \infty}\Big|\int_0^1kx^kf(x)\,dx - a\Big| \le \epsilon.$To conclude let $\epsilon \to 0^+$.