Show that $f(x)=\frac{1}{\sqrt{x}}$ is uniformly continuous on the domain $(1,\infty)$ but not on the domain $(0,1)$.

Question

Show that $f(x)=\frac{1}{\sqrt{x}}$ is uniformly continuous on the domain $(1,\infty)$ but not on the domain $(0,1)$.

$\def\verts#1{\left\vert#1\right\vert}$ My Attempt

First we show that $\forall\varepsilon>0,\exists\delta>0$ such that $\forall x,y\in (1,\infty)$, if $\verts{x-y}<\delta$ then $\verts{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}}<\varepsilon$

Since $x,y\in(1,\infty)$, that $0<\frac{1}{\sqrt{x}},\frac{1}{\sqrt{y}}<1$ i.e. $\verts{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}}<1$, if $1<|x-y|$, let $\delta=\varepsilon$ then $\verts{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}}<\varepsilon$. If $\verts{x-y}\le1$ then $\dots$ here i'm stucked on the second case

Next is to show $\exists\varepsilon>0$,$\forall\delta>0,\exists x,y\in(0,1)$ such that $\verts{x-y}<\delta$ and $\verts{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}}\ge\varepsilon$, not sure how to prove this, could someone help me.

Answer 1

As in the hint from @BrianMoehring, if $n^2>1/\varepsilon$ then $|1/n^2-1/(n+1)^2|<1/n^2<\varepsilon$ and $|\sqrt {1/(1/n^2)}-\sqrt {1/(1/(n+1)^2)}|=1.$

If $a<b$ and $f:(a,b)\to \Bbb R$ is uniformly continuous on $(a,b)$ then $f$ is bounded on $(a,b).$

For we may take $r>0$ such that $|f(x)-f(y)|<1$ whenever $x,y\in (a,b)$ with $|x-y|<r.$ Now take some (any) $x_0\in (a,b).$ Consider some $n_0\in \Bbb N$ such that $x_0-n_0r/2\le a$ and $x_0+n_0r/2\ge b.$

Then $|f(x)|<n_0+|f(x_0)|$ for all $x\in (a,b).$

E.g. if $n_0> 2$ and $x_0+(r/2)<x\le x_0+2(r/2)<b$ then $$|f(x)|\le$$ $$\le |f(x)-f(x-[x_0+2(r/2)])|+{}$$ $${}+|f(x_0+2(r/2))-f(x_0+r/2)|+{}$$ $${}+|f(x_0+r/2)-f(x_0)|+{}$$ $${}+|f(x_0)|<$$ $$<3+|f(x_0)|\le n_0+|f(x_0)|.$$

So if $f$ is unbounded on $(a,b),$ e.g. if $f(x)=1/\sqrt x$ and $(a,b)=(0,1),$ then $f$ cannot be uniformly continuous on $(a,b).$

Answer 2

One side is easy: $f(x)$ is unbounded on $(0, 1)$ so it can’t be uniformly continuous there. [For any small positive $\delta$, partition $(0, 1)$ into finitely many intervals of length at most $\delta$; $f(x)$ is unbounded on the whole interval, so it must be unbounded on one of these subintervals.]

On the other side, we can use Mean Value Theorem and the fact that $|f’(x)| = |x^{-3/2}/2| \leq 1/2$ on $(1, \infty)$. Mean Value Theorem guarantees that for any $1 < x < y < \infty$ there exists $c \in [x, y]$ such that $$|f(x) -f(y)| = |f’(c)| |x - y| \leq |x - y|/2,$$ so for any $\epsilon$ the corresponding $\delta = 2\epsilon$ works for all $x, y \in (1, \infty)$.

Answer 3

For a function $f : E \to \mathbb{R}$ defined on a subset $E$ of $\mathbb{R}$, the uniform continuity is equivalent to what is called the microcontinuity. This property can be stated by two equivalent ways (with the second version requiring basic knowledge on hyerreal numbers):

1. For any two sequences $(a_n)$, $(b_n)$ in $E$ such that $|a_n - b_n| \to 0$, we have $|f(a_n) - f(b_n)| \to 0$.

2. Let $f^*$ be the natural extension of $f$ in the field $\mathbb{R}^*$ of hyperreal numbers. Then $f^*(x) \approx f^*(y)$ (i.e., they are infinitely close) for any $x, y \in E^*$ with $x \approx y$.

The equivalence of the uniform convergence and the first version of mircocontinuity can be proved in a fairly routine way, and so, I will omit the proof. (Note: This may be viwed as an analogue of the equivalence of two definitions of continuity: $\epsilon$-$\delta$ version and the sequential version.)

This property often works better when proving that a given function is not uniformly continuous. For instance, let $f : (0, \infty) \to \mathbb{R}$ be defined by $f(x) = 1/\sqrt{x}$. Then the choices

$$ a_n = \frac{1}{n^2}, \qquad b_n = \frac{1}{(n+1)^2} $$

satisfy $|a_n - b_n| \to 0$ but $|f(a_n) - f(b_n)| = 1 \not\to 0$. Therefore $f$ is not uniformly continuous.