True of false: $|f(x)-g(x)|<\epsilon$ $\forall x\in I$ $\Rightarrow$ $|\sup(f(x))-\sup(g(x))|\leq \epsilon$

Question

True of false: $|f(x)-g(x)|<\epsilon$ $\forall x\in I$ $\Rightarrow$ $|\sup(f(x))-\sup(g(x))|\leq \epsilon$

I actually have an idea for a proof in case it is correct, but just to make sure what I'll write isn't rubbish I would like to get a confirmation from you all.

Intuitively, it seems logical to me.

If anyone has a counter example I would love to see it.

Thanks a lot.

Answer 1

Let $a=\sup (f(x))$ and $b=\sup(g(x))$. Then, for all $\delta>0$, we have that there is some $x_0\in I$ such that $ f(x_0)\geq a-\delta$. Since $|f(x_0)-g(x_0)|<\epsilon$. Then, $$a-\delta-\epsilon\leq f(x_0)-\epsilon< g(x_0)\leq b$$ So, $a-\delta-\epsilon< b$. Since this inequality works for any $\delta>0$, by taking $\delta\to 0^+$ we conclude $a-\epsilon\leq b$.

Similarly, $b-\epsilon\leq a$. So, we conclude $|b-a|\leq \epsilon$.

Answer 2

A simple explanation :

If the distance between any two given values of $f(x)$ and $g(x)$ is smaller than $\epsilon$, then the distance between their supremums can follow as such as well. Of course, that only fails in the case of either one (or both) of the supremums being infinite.

So, yes, what you write isn't wrong. You may continue with your desired statement/proof.