A calculus problem regarding continuity and uniform continuity

Question

A function $f : \mathbb{R} \to \mathbb{R} $ is continuous at the origin $x=0$ and we have

$$ f(0) = 0 $$

$$ \forall x_1 \in \mathbb{R} ,\forall x_2 \in \mathbb{R}: f(x_1 + x_2 ) \leq f(x_1) + f(x_2) $$

does it follow that $f$ is uniformly continuous?

Answer 1

We have $f(x+h)\leq f(x) + f(h)$, so$$f(x+h) - f(x) \leq f(h)$$

Similarly we have $f(x) =f(x+h -h)\leq f(x+ h) + f(-h)$, so$$f(x+h) - f(x) \geq -f(-h)$$

Now we have $$-f(-h)\leq f(x+h) - f(x) \leq f(h)$$

so $$|f(x+h) - f(x)| \leq \max\{|f(h)|, |f(-h)|\}$$

Now use the continuity at $x=0$ to conclude:

$f(0) =0$ and $f$ is continuous at $x=0$ means for any $\epsilon >0$, $\exists \delta >0$ such that for all $|h| < \delta$, $|f(h)|= |f(h) - f(0)|< \epsilon$

So once $|h| < \delta$, we have $|f(h)|< \epsilon$ and $|f(-h)|< \epsilon$, so

$$|f(x+h) - f(x)| \leq \max\{|f(h)|, |f(-h)|\} < \epsilon, \forall x\in \mathbb{R}$$

So $f$ is uniformly continuous.