Proving $\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$ when $f$ is continuous on $[0,1]$.

Question

Show $$\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$$ when $f$ is continuous on $[0,1]$.

I know it can be proved using the bounded convergence theorem but I want to know a proof using only basic properties of the Riemann integral, the fundamental theorem of calculus, and the mean value theorem for integrals. Thank you.

Answer 1

take any $\epsilon$, choose $\delta > 0$ such that $|f(x) - f(0)| < \epsilon$ on $[0,\delta]$. choose $n$ big enough such that $(1-\epsilon)^n < \delta$ then $$ |\int_0^1 f(x^n) dx - f(0) |= |\int_0^{(1-\epsilon)} [f(x^n) - f(0)]|dx + \int_{1-\epsilon}^1 [f(x^n) - f(0) ]dx | \leq $$ $$ \int_0^{(1-\epsilon)} |f(x^n) - f(0)|dx + \int_{1-\epsilon}^1 |f(x^n) - f(0) |dx $$ first factor is smaller than $\epsilon(1 - \epsilon)$ thanks to the choice of $\delta$ and $n$, second one is smaller than $\epsilon \cdot 2 \sup |f|$ because length of your interval of integration is $\epsilon$ so the result follows since $\epsilon$ was arbitrarily small

Answer 2

Hint:

$f(0)=\int_0^1 f(0) dx$

$x^n\to 0$ for $0\leq x<1$

and $f$ is continuous.

Answer 3

$\int_0^{\infty}(f(e^{-nt})-f(0))e^{-t}dt\to 0$ by dominated convergence, since $f$ is contimuous, therefore bounded, and since $\int_0^{\infty}e^{-t}dt$ converges.

Answer 4

Handsome tip:

Prove it for $g\left(x\right)=f\left(x\right)-f\left(0\right)$ where $g\left(0\right)=0$ and then make use of $\int_{0}^{1}f\left(x^{n}\right)dx=\int_{0}^{1}g\left(x^{n}\right)+f\left(0\right)dx=\int_{0}^{1}g\left(x^{n}\right)dx+f\left(0\right)$.