Prove that $f(x)=x^2$ is not uniformly continuous on $[a,\infty).$

Question

Prove that $f(x)=x^2$ is not uniformly continuous on $[a,\infty).$

The solution given is as follows:

Let us choose $\epsilon > 0.$ Then for any two points $x,c\in [a,b]$ satisfying $|x-c|< \delta,$ the inequality $| f(x) - f(c)|< \epsilon$ will hold if we choose $\delta= \epsilon/2b.$

But as $b$ takes larger and larger values, $\delta$ gets smaller and smaller. So it is not possible to find a single positive $\delta$ which will work for all $b > a.$ It follows that $f$ is not unifOrmly continuous on $[a,\infty).$

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However, I don't think this solution works. This is because, as the solution asserts:

But as $b$ takes larger and larger values, $\delta$ gets smaller and smaller. So it is not possible to find a single positive $\delta$ which will work for all $b > a$

I think, this needs a little more justification as, maybe the scheme of choosing a delta by the formula $\delta=\epsilon/2b$ does not work but it does not seem to suggest in any way that no scheme of choosing $\delta\gt 0$ varying with only $\epsilon$ works.

I think, a proper solution will require us to show one of the following things as a non-uniformity criteria :

Let $f:A\to \Bbb R$ and $A\subseteq \Bbb R$ and $f$ is not uniformly continuous on $A$ iff either of the following twp conditions are satisfied:

• There exists an $\epsilon _0> 0$ such that for every $\delta > 0$ there are points $x_{\delta}, c_{\delta}$ in $A$ such that $x_{\delta}- c_{\delta}\lt \delta$ and $f(x_{\delta})-f( c_{\delta}) \geq \epsilon_0.$

• There exists an $\epsilon_0 > 0$ and two sequences $(x_n)$ and $(u_n)$ in $A$ such that $\lim (x_n -u_n)= 0$ and $|f(x_n)-f(u_n)|\geq \epsilon_0$ for all $n\in \Bbb N.$

Or maybe we can use the contrapositive of the statement (below):

If $f : A\to \Bbb R$ is uniformly continuous on a subset $A$ of $\Bbb R$ and if $(x_n)$ is a Cauchy sequence in $A,$ then $(f(x_n))$ is a Cauchy sequence in $\Bbb R$

to establish our claim.

Answer 1

Let's prove it by contraposition. Take $\varepsilon>0$, and let $\delta>0$ be arbitrary. Observe that for any $x,y\in[a,\infty)$ we have that

$$\lvert f(x)-f(y)\rvert=\lvert x^2-y^2\rvert=\lvert x-y\rvert\lvert x+y\rvert.$$

In particular for any $x\in[a,\infty)$, if we set $y=x+\frac{\delta}{2}$, then $\lvert x-y\rvert=\frac{\delta}{2}<\delta$, but

$$\lvert f(x)-f(y)\rvert=\frac{\delta}{2}\left\lvert 2x+\frac{\delta}{2}\right\rvert.$$

In particular, if we take

$$x=\max\left\{0,a,\frac{\varepsilon}{\delta}-\frac{\delta}{4}\right\},$$

then

$$\lvert f(x)-f(y)\rvert\geq\frac{\delta}{2}\left(2\left(\frac{2\varepsilon}{\delta}-\frac{\delta}{4}\right)+\frac{\delta}{2}\right)=\varepsilon.$$

This proves that $f$ is not uniformly continuous on $[a,\infty)$.

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Here's another solution using the second sequential characterization of uniform continuity you gave. Consider the sequences $\{x_j\}_{j\in\mathbb{Z}^+}$ and $\{y_j\}_{j\in\mathbb{Z}^+}$ in $[a,\infty)$ defined by

$$x_j=a+j^2,\quad y_j=a+j^2+\frac{1}{j}$$

for all $j\in\mathbb{Z}^+$. Then

$$\lim_{j\to\infty}(x_j-y_j)=\lim_{j\to\infty}\frac{1}{j}=0,$$

but

$$\lvert f(x_j)-f(y_j)\rvert=\frac{2a+2j^2+\frac{1}{j}}{j}=2j+\frac{2a}{j}+\frac{1}{j^2}\geq 2$$

for all $j\in\mathbb{Z}^+$.

Answer 2

Here is an alternative proof, which is somewhat simpler.

Suppose that $f$ is uniformly continuous.

Let $\varepsilon >0$ be arbitrary and pick $\delta >0$ be s.t. $|x-y| < \delta \implies |f(x) - f(y)| < \varepsilon$.

Note that by the mean value theorem, for (WLOG) $x < y$,

$$f(y)-f(x) = f'(c) (y-x)$$

$$\implies f'(c) (y-x) < \varepsilon$$

for some $c \in (x,y)$.

If we fix $y = x +\frac{\delta}{2}$, then by the uniform continuity condition, we have that, for every $x$,

$$\frac{\delta}{2} f'(c) < \varepsilon$$

But $x \to \infty \implies f'(c) \to \infty$, so this leads to a contradiction. $\square$

It is not true in general that uniformly continuous $\implies$ bounded derivative, but it is often worth a shot.