Uniform convergence of $f_n(x)=\int_0^x f(t^n)dt$, where $f$ is continuous on $[0,1]$.

Question

How to prove uniform convergence of $f_n(x)=\int_0^x f(t^n)dt$ on $[0,1]$, where $f$ is continuous on $[0,1]$.

My attempt. By changing variables $t^n=s$, we see $|f_n(x)|\leq x\max_{[0,x^n]}|f|$. This shows that $f_n(x)\to 0, x\in [0,1)?$. Should we need an assumption: $f(0)=0$? Then what to do?

Let $f(t)=t$, we see $f_n(x)=x$, contradicting to the above estimate. Oh. What's wrong?

I have noticed the problem have been asked before, Studying the pointwise and uniform convergence of $f_n(t)= \int^{t}_{0}f(x^n)dx$ but with no answer.

$f(t) = t$ gives $f_n(x) = \int_0^x t^n dt = x^{n+1}/(n+1)$. — Martin R, Jan 04 '23 at 12:26
And why should $x\max_{[0,x^n]}|f|$ converge to zero for $n \to \infty$? Or are you assuming that $f(0) = 0$? — Martin R, Jan 04 '23 at 12:29

score 1 · Accepted Answer · answered Jan 04 '23 at 12:32

Hint: One expects the limit to be $f(0)x$ since $f(t^{n})\to f(0)$ for all $t<1$. Now consider $\int_0^{x} [f(t^{n})-f(0)] dt$. First note that $f$ is bounded so $\int_s^{1}|f(t^{n})-f(0)|dt <\epsilon$ for all $n$ is $s$ is chosen sufficiently close to $1$. Now use continuity of $f$ at $0$ to show that $\int_0^{x}|f(t^{n})-f(0)|dt <\epsilon$ for all $x \leq s$ provided $n$ is large enough. Hence, the convergence is uniform on $[0,1]$.

Uniform convergence of $f_n(x)=\int_0^x f(t^n)dt$, where $f$ is continuous on $[0,1]$.

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