Prove that $f(x)=1/\sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.
I believe it follows that $f(x)$ is not uniformly continuous because $f(x)$ is not continuous on the bounded interval including zero. I tried to set up an $\epsilon-\delta$ proof of the continuity on $(0,1]$ with $\delta=\frac{p}{(2\epsilon+1)^2}$ where $p$ is any point in the interval, but I feel like I am making a leap to say that by that $\delta$ we get $|\frac{\sqrt p-\sqrt x}{\sqrt p\sqrt x}|\lt\epsilon$
Best Answer
Recall that $f$ is uniformly continuous if and only if for each $\varepsilon>0$, there exists $\delta>0$ such that $$\sup\{|f(x)-f(y)| : |x-y|<\delta\}<\varepsilon.$$ Consider the sequence $x_n=\frac1{n^2}$. Then for any $N$,
$$ |x_N-x_{N+m}| = \frac1{N^2}-\frac1{(N+m)^2} \stackrel{m\to\infty}\longrightarrow\frac1{N^2}, $$ and $$|f(x_N)-f(x_{N+m})|=\left|\frac1N-\frac1{N+m} \right|=\frac{m}{N(N+m)}. $$ Choose $\varepsilon=\frac12$. Then for any $\delta>0$, we may choose $N$ such that $\frac1{N^2}<\delta$, but $$f(x_N)-f(x_{N+m}) = \frac{m}{N(N+m)}> \frac m{\delta+m}\stackrel{m\to\infty}\longrightarrow 1,$$ so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>\frac12.$$ It follows that $f$ is not uniformly continuous