2
$\begingroup$

If $f: [0, \infty[ \to \mathbb{R}^p$ is continuous and satisfies $\lim_{x \to \infty} f(x + y) - f(x) = 0$ for all $y \in [0, 1]$, then $f$ is uniformly continuous.

Attempt: suppose $f$ is not uniformly continuous. Then $\exists \epsilon' > 0, \forall M > 0, \exists x > M, \exists y \in [0, 1]$ such that $$\|f(x + y) - f(x)\| \geq \epsilon'.$$ We'll call those $x, y$ $x', y'$ respectively. I'm trying to show that $f$ cannot be continuous in this case. So I need to find an $a$, such that for all $\delta > 0$, $$|a - b| < \delta \implies \|f(a) - f(b)\| \geq \epsilon$$ for some $\epsilon$. Now that $\epsilon$ is bound to be $\epsilon'$, but I'm stuck on finding such $a$ and $b$, because $|x + y - x| = y$ which is always greater than an $\delta > 0$ unless $y = 0$, but $y$ is not necessarily $0$. We can make $x$ arbitrarily large by increasing $M$, but I don't see how that could help finding $a, b$ that are close to each other. Am I overlooking something?

$\endgroup$
3
$\begingroup$

Hint. If is $f$ is continuous on $\mathbb{R}$, but not uniformly continuous then there exists $\epsilon>0$, and there are two sequences $(x_n)_n$, $(y_n)_n$ such that $x_n\to+\infty$, $y_n\to+\infty$, $x_n-y_n\to 0$ and $\|f(x_n)-f(y_n)\|>\epsilon$.

$\endgroup$
  • $\begingroup$ Thanks for your hint. I now have a proof, but it doesn't use that $f$ is continious. Is it correct? Proof: we keep $x', y', \epsilon'$ defined the same way as in the original post. Then by the assumption from the title, for $y'$, there exists an $M'$ such that for all $x > M', ||f(x + y') - f(x)|| < \epsilon'$. But that contradicts the assumption that for all $M$, so $M'$ in particular, there exist an $x' > M$ such that $ ||f(x + y') - f(x)|| \geq \epsilon'$ $\endgroup$ – Pel de Pinda Nov 7 '17 at 15:43
  • $\begingroup$ @PeldePinda That's correct. The fact that $f$ is continuous is important. It forces the two sequences $(x_n)_n$, $(y_n)_n$ to be unbounded (otherwise u.c. on compact sets is contradicted). $\endgroup$ – Robert Z Nov 7 '17 at 16:04
  • $\begingroup$ But the proof I wrote down doesn't use that $f$ is continious, so then it must be wrong right? $\endgroup$ – Pel de Pinda Nov 7 '17 at 16:11
  • 1
    $\begingroup$ If $f$ is not continuous in $[-1,1]$, but u.c. outside it, then your your proof does not work! $\endgroup$ – Robert Z Nov 7 '17 at 16:17

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.