How to prove that g is not uniformly continuous?

Question

Let the function $g : ]0,1] \rightarrow\mathbb{R}$ be given by

$$ g(t) = \frac{1}{t} $$

I then have to show that g is continuous but not uniformly continous. To prove that g is continuous I have done the following:

Let $a \in ]0,1]$. Then we have that $|g(a) - g(t)| = |\frac{1}{a} - \frac{1}{t}| = |\frac{t-a}{at}| = |t-a| \cdot \frac{1}{at}$. Setting $ \delta = \min\{\frac{a}{2},\epsilon \frac{a^2}{2}\}$ we have $|t-a| \cdot \frac{1}{at} \leq \epsilon \frac{a^2}{2} < \delta $ and we are done. Is this correct?

And how do I prove that g is not uniformly continuous? Thanks for your help.

Answer 1

$g$ is uniformly continuous if and only if $$ \forall \epsilon>0 \ \ \ \ \exists\delta>0 \ \ \ \ s.t \ \ \ |x-y|<\delta\ \ \Rightarrow\ \ \ |g(x)-g(y)|<\epsilon $$

So $g$ is not uniformly continuous if and only if

$$ \exists\epsilon>0\ \ \ s.t\ \ \forall \delta>0 \ \ \ \exists x,y \ \ , |x-y|<\delta \ \ , |f(x)-f(y)|\geq \epsilon $$

We can define $\epsilon =1$. Then:

$$ \epsilon=1 \ \ \ \ \ \ \text{so} \ \text{if}\ \ \ 1> \delta >0 \ \ \ \ \ x:=\frac{\delta}{2}, y=\frac{\delta}{3 } \ \ \Rightarrow \ |x-y|=\frac{\delta }{6}<\delta \ \ But \ \ |f(x)-f(y)|=\frac{1}{\delta }>1 $$

Answer 2

You can prove it by contradiction. Suppose you have a universal $\delta$ for some fixed $\epsilon>0$. From your calculations you know that $$ |g(t)-g(a)|=|t-a|/at $$ If you choose $a<\min\{\delta,1/2\epsilon\}$ and $t=a/2$ (thus $|t-a|<\delta$), you get $$ |g(t)-g(a)|=1/a>2\epsilon, $$ a contradiction

Answer 3

Your $\epsilon-\delta$ proof is not correct, but it is not needed because it suffices to note that $f(t) = t$ is everywhere continuous and nonvanishing on $(0,1]$, hence $g(t) = \frac1 {f(t)}$ is continuous on $(0,1]$.

If you do want an $\epsilon-\delta$ proof, then given $a$ and $\epsilon > 0$, we want to find $\delta$ small enough such that

$$0<|t-a|<\delta \implies |g(a)-g(t)| = \frac{|t-a|}{at} < \frac{\delta}{at} < \epsilon.$$

We have $0<|t-a|<\delta \implies t > a - \delta$ so if $\delta < a$, this lower bound will be positive. Then:

$$\frac\delta{at} < \frac\delta{a(a-\delta)}$$

and this expression does not involve $t$. So we wish to find $0<\delta<a$ such that

$$\frac\delta{a(a-\delta)}< \epsilon \iff \delta < \epsilon a^2 - \epsilon a \delta \iff \delta(1+\epsilon a) < \epsilon a^2 \iff \delta< \frac{\epsilon a^2}{1+\epsilon a}.$$

Therefore, we can pick

$$\delta < \min\left\{a, \frac{\epsilon a^2}{1+\epsilon a}\right\}.$$

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To show that $g$ is not uniformly continuous, let $\epsilon > 0$ and $\delta > 0$. Notice that if $a \in(0,1-\delta/2]$ and $t=a+\delta/2$, then $|t-a| = \delta/2$, and hence

$$|g(a) - g(t)| = \frac\delta{2at}.$$

Letting $a\to 0$ in the expression above, we see that $|g(a)-g(t)|$ is unbounded, and in particular $>\epsilon$, which violates the requirement for uniform continuity.

What's important here is that $a$ and $t$ are free to vary after $\epsilon$ and $\delta$ are chosen.