1
$\begingroup$

True or False: If $f \colon [a, \infty) \rightarrow \mathbb{R}$ is bounded and continuous, then $f$ is uniformly continuous on $[a, \infty)$.

If true, supply a proof. If false, provide a counter example.

So my thinking is that it is true. I couldn't think of a counter example for this so that's why I believe it is true. What I am having trouble with is showing that it is true. I know the definition for $f$ to be continuous is: $\forall \epsilon >0, \exists \delta > 0,$ such that if $x \in [a, \infty$) and $|x-c|< \delta$ then $|f(x)-f(c)|< \delta$.

I know the definition for $f$ to be uniformly continious is $\forall \epsilon >0, \exists \delta > 0,$ such that if $x,y \in [a, \infty$) nd $|x-y|< \delta$ then $|f(x)-f(y)|< \delta$.

I was sort of thinking about "combining" these two definitions, however the $\delta$ in the first definition is dependent on $c$ and the $\delta$ in the second definition CANT depend on $c$ (in this case $y$).

Is this true? can someone hep me prove this

$\endgroup$
1

1 Answer 1

2
$\begingroup$

Consider $f(x)=\sin x^2$ .

Then $ x_n=\sqrt{\dfrac{(n+1)\pi}{2}}$ and $y_n=\sqrt{\dfrac{n\pi}{2}}$

Choose $\delta$ such that $|x_n-y_n|<\delta $ for $n $ large but we have $|\sin(x_n)-\sin (y_n)|=1$

$\endgroup$

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.