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?


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 $\|f(x_n)-f(y_n)\|>\epsilon$.

  • $\begingroup$ 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
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    $\begingroup$ 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

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