Show $|f(x + \delta_n) - f(x)| < \epsilon_n$ for $\epsilon_n$ independent of $x$

Question

For a sequence $\delta_n \rightarrow 0$ I need to show that there exists another sequence $\epsilon_n \rightarrow 0$ such that:

$$\sup_x|f(x + \delta_n) - f(x)| < \epsilon_n$$

What is required for this property to hold? Does it have a name?

If this did not have the supremum over $x$, I can find such an $\epsilon_n$ if the left and right limits are equal at all $x$. Is it sufficient that $f$ is bounded on it's domain? Is that too strong?

Answer 1

Here is why uniform continuity is enough:

Let $\epsilon_n = 2 \sup | f (x+\delta_n) - f x |$.

This obviously satisfies the first condition you want. So we just need to show that $\epsilon_n \to 0$.

Let $\epsilon^* > 0$. By uniform continuity, we can find a $\delta^*$ so that whenever $\delta_n < \delta^*$, $\sup | f(x+\delta_n) - f x | < \epsilon^*$.

Then since $\delta_n \to 0$, we know $\delta_n < \delta^*$ for $n$ large enough.

But this means, for $n$ large enough, $\epsilon_n = 2 \sup | f(x+\delta_n) - f x | < 2 \epsilon^*$.

So $\epsilon_n \to 0$.

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I hope this helps ^_^