Show that $h_n(x)=x^{1+frac{1}{2n-1}}$ converges uniformly on $[-1, 1]$.

Question

This is a question about the uniform convergence of a sequence of functions.
First, note that $lim_{ntoinfty}h_n=|x|$ is known already. We simply need to get an appropriate bound on $|h_n(x)-h(x)|$.
Here's my attempt (note crucially that the domain is $[-1, 1]$ instead of $mathbb{R}$):
$$
|h_n(x)-h(x)|=|x^{1+frac{1}{2n-1}}-|x||=|xcdot x^{frac{1}{2n-1}}-|x||leq|x||x^{frac{1}{2n-1}}|+|x|=(|x^{frac{1}{2n-1}}|+1)|x|leq |x^{frac{1}{2n-1}}|+1leq2.
$$
which obviously isn't helpful because the final expression does not involve $n$. In other words, I can't choose $n$ in response to some $epsilon$-bound on the error $|h_n(x)-h(x)|$.
What are the required modifications? Your help is much appreciated.

Kavi Rama Murthy · Answer

It is enough to show that $h_n(x) to x$ uniformly on $[0,1]$ since $h_n$ is even. Let $epsilon >0$ and note that $|h_n(x)-x| leq |x||x^{frac 1 {2n-1} }-1|leq 2x<epsilon $ if $x <frac {epsilon} 2$. Now let $x geq frac {epsilon} 2$. Then $|h_n(x)-x| leq |x^{frac 1 {2n-1} }-1|=|e^{frac 1 {2n-1} ln x }-1|$. So $|h_n(x)-1| leq e^{t_n} frac 1 {2n-1} |ln x|$ for some $t_n$ between $0$ and $frac 1{2n-1} ln x$ by MVT applied to the exponential function and observe that $ln x$ is bounded for $geq frac {epsilon} 2$. Can you finish the proof?

Answered by Kavi Rama Murthy on January 17, 2021

Show that $h_n(x)=x^{1+frac{1}{2n-1}}$ converges uniformly on $[-1, 1]$.

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