0
$\begingroup$

I have to prove that the function $f(x)=\frac{1}{x}$ on $(0,\infty)$ is not uniformly continuous (for the definition of uniform continuity see here).

I negated the definition of uniform continuity getting:

$$ \exists \epsilon>0 \ \text{s.t.} \ \forall \delta>0 \ \text{I can always find} \ x,y \in(0,\infty) \ \text{s.t.} \mid x-y\mid<\delta \ \text{but} \ \mid f(x)-f(y)\mid \geq \epsilon. $$

So I chose $\epsilon=1$. Set $y=\frac{x}{2}$. Then I have to find an $x$ such that this holds: $\mid x-y\mid = \frac{x}{2} <\delta$ & $\mid f(x)-f(y)\mid = \frac{1}{x} \geq 1$, which is equivalent to $ x<\min [2\delta,1] $.

Is this argument valid? My concern is if I'm allowed to choose $x$ depending on $\delta$. Thanks!

$\endgroup$
0
$\begingroup$

It is correct. You can choose x depending on $\delta$ because $x$ is quantified before $\delta$.

Alternatively, we have $$1/n - 1/(n+1) \to 0$$ but

$$f(1/n) -f(1/(n+1)) = n - (n+1) \to -1 $$

And by the equivalent sequence characterisation of uniform continuity, we conclude that $f$ cannot be uniformly continuous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.