Convergence of $h_{n}(x) = x^{ 1 +\frac{1}{2n-1} }$ defined on [-1,1]

Question

1. $$ h_{n}(x) = x^{ 1 + \frac{1}{2n-1} } = x^{ \frac{2n-1+1}{2n-1} } = x^{\frac{2n}{2n-1}} = (x^2)^{\frac{n}{2n-1}} = (x^2 )^{ \frac{1}{2-\frac{1}{n}} } $$

then $$ \lim_{n\rightarrow \infty} h_{n}(x) = (x^2)^{\frac{1}{2}} = |x| $$

2. $ \ \ $ I want to know if the convergence is pointwise or uniform.

$\ \ \ \ \ $ Let $x \in [-1,1]$

$$ | h_{n}(x)-|x| | = | x^{ 1 + \frac{1}{2n-1} } - |x|| = | x^{1}\ x^{\frac{1}{2n-1}} - |x| | = | x\big( x^{\frac{1}{2n-1}}-(\pm1) \big) | \leq |x^{\frac{1}{2n-1}}-(\pm1) \big)| $$ since $|x|\leq 1 \ \Big(x\in [-1,1]\Big)$

but I'm stuck. How would you do that?

Answer 1

The convergence is indeed uniform. Note that $h_n$ and $h(x)=|x|$ are both even functions so it is enough to prove uniform convergence on $[0,1]$. So consider $|x^{1+\frac 1 {2n-1}} -x|$ with $0\leq x \leq 1$. This is $x|1-e^{\frac 1 {2n-1} \log x}|$. Use the inequality $1-e^{-t} \leq t$ for all $t \geq 0$ to see that $x|1-e^{\frac 1 {2n-1} \log x}| \leq \frac {(-x\log x)} {2n-1}$. Note that $x\log x$ is bounded, so the last quantity tends to $0$ uniformly.