Given a continuous function $f:[0,1] \to R$, prove that $\lim_{n\to\infty}\int_0^1f(x^n)dx = f(0)$.
Attempt:
Let $u = x^n$, so $du = nx^{n-1}dx$. Then substituing $u$ in I got:
$\lim_{n\to\infty}\dfrac{\int_0^1f(u)du}{nx^{n-1}}$
Doesn't this limit go to $0$? I'm not sure which part I'm messing up on, any hints are appreciated.
It's equivalent to show $\lim_{n\rightarrow\infty}\int_0^1 (f(x^n)-f(0))dx=0$.
Given any $\epsilon>0$, we have $$\left|\int_0^1 (f(x^n)-f(0))dx\right|\le\int_0^1 |f(x^n)-f(0)|dx$$ $$=\int_0^{1-\epsilon}|f(x^n)-f(0)|dx + \int_{1-\epsilon}^1 |f(x^n)-f(0)|dx$$ $$\le\int_0^{1-\epsilon}|f(x^n)-f(0)|dx + 2B\epsilon$$
where $B=\max_{x\in[0,1]}|f(x)|$ which exists due to continuity.
By $f(x)$ is continuous at $x=0$, there is a $\delta>0$, such that $|x|\le\delta\Rightarrow |f(x)-f(0)|\le \epsilon$. Thus when $n\ge \log_{1-\epsilon}\delta$, $|x^n|\le (1-\epsilon)^n\le\delta$ for $x\in [0, 1-\epsilon]$, and for sufficiently large $n$, $$\int_0^{1-\epsilon} |f(x^n)-f(0)|dx\le \int_0^1 \epsilon dx \le \epsilon$$
Finally,
$$\left|\int_0^1 (f(x^n)-f(0))dx\right|\le (1+2B)\epsilon$$
In fact, we only need $f(x)$ is bounded and right continuous at $0$ (and integrable).
$\newcommand{\d}{\ {\mathrm d}} \newcommand{\e}{\varepsilon}\newcommand{\md}[1]{\left\lvert #1 \right\rvert}$Here's a bare-hands elementary proof without any appeal to the Dominated Convergence Theorem.
Note that we have $$\int_0^1 f(x^n) \d x - f(0) = \int_0^1 [f(x^n) - f(0)] \d x.$$
Let $M := \sup_{[0, 1]} |f|$. Note that this is finite since $f$ is continuous.
Let $\e > 0$ be given. Fix $\eta \in (0, 1)$ such that $2 (1 - \eta) M < \frac{\e}{2}$.
By continuity at $0$, there exists $\delta > 0$ such that $$|y| < \delta \Rightarrow |f(y) - f(0)| < \frac{\e}{2}.$$
Note that $x \mapsto \md{x^n}$ is increasing on $[0, \eta]$ and $\eta^n \to 0$. Thus, there exists $N \in \Bbb N$ such that $$|x^n| < \delta \text{ for all } x \in [0, \eta] \text{ and all } n \geqslant N.$$
Thus, for $n \geqslant N$, we have
\begin{align} \md{\int_0^1 [f(x^n) - f(0)] \d x} & \leqslant \int_0^1 \md{f(x^n) - f(0)} \d x \\ &= \int_0^{\eta} \md{f(x^n) - f(0)} \d x + \int_{\eta}^{1} \md{f(x^n) - f(0)} \d x \\ &\leqslant \int_0^{\eta} \frac\e2 \d x + \int_{\eta}^{1} 2M \d x \\ &\leqslant \frac\e2 \eta + 2M (1 - \eta) \\ &< \e. \end{align}
Without loss of generality, $f(0)=0.$ Define $F_n(x)=\int_0^xf(t^n)dt.$
Let $\epsilon>0$ and choose $0<\delta<1$ so that
$\tag1 0<x<\delta \Rightarrow |f(x)|<\epsilon$ and write
$\tag2 \int_0^1f(t^n)dt=\int_0^af(t^n)dt+\int_a^1f(t^n)dt.$
Apply the mean value theorem to find a $c_n\in (a,1)$ such that
$\tag3 F_n(1)-F_n(a)=\int_a^1f(t^n)dt=(1-a)f(c_n^n).$
Note that $f$ is bounded on $[0,1]$ so
$\tag4 (1-a)|f(c_n^n)|<M(1-a)\ \text{for some}\ M>0.$
Choose $a$ so that $M(1-a)<\epsilon$ and then choose $N$ large enough so that $a^N<\delta.$ Then, $t^n<\delta$ whenever $0<t<a$ and $n\ge N.$
Then, from $(1),(2), (3)$ and $(4),$
$\tag5 \left |\int_0^1f(t^n)dt\right|\le \left |\int_0^af(t^n)dt\right |+\left |\int_a^1f(t^n)dt\right |<\epsilon+ \epsilon=2\epsilon.$
