A function which is continuous at $x=0$ and seems to be uniform continuity.

Question

Given $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous at zero, $f(0)=0$ and satisfies: $f(x+y)\leq f(x)+f(y),\forall x,y\in \mathbb{R}$

Prove that exist $\delta >0$ such that with any $x,y\in \mathbb{R}$ that $\left | x-y \right |<\delta$, we always have: $\left | f(x)-f(y) \right |<\frac{1}{10000}$

I have read the definition of uniform continuity: for all ${\displaystyle \epsilon >0}$ there exists a ${\delta >0}$ such that for all ${x,y\in \mathbb{R},|x-y|<\delta \implies |f(x)-f(y)|<\epsilon }$. It is really similar to my problem.

I have a trial: choose any real number $a$, from $f(x-a)+f(a) \geq f(x)$, then $\lim\limits_{x \to a}f(x-a)+f(a)\geqslant \lim\limits_{x \to a}f(x)\Rightarrow f(a)\geq \lim\limits_{x \to a}f(x)$, from $f(a-x)+f(x) \geq f(a)$, we also have $f(a)\leq \lim\limits_{x \to a}f(x)$ which implies $f(a)=\lim\limits_{x \to a}f(x)$ so $f$ is continuous at $x=a$, then $f$ is continuous on $\mathbb{R}$.

Is my solution wrong? If it is fine,then what's the next thing to do?

Answer 1

Since $f$ is continuous at $0$ and $f(0)=0$, there is a number $\delta\gt0$ such that $$|u|\lt\delta\implies f(u)\lt\frac1{10000}$$ Now suppose $x,y\in\mathbb R$ and $|x-y|\lt\delta$. Without loss of generality suppose $f(x)\ge f(y)$. Then $$f(y)\le f(x)=f(y+(x-y))\le f(y)+f(x-y)$$ so $$0\le f(x)-f(y)\le f(x-y)\lt\frac1{10000}$$ so $$|f(x)-f(y)|\lt\frac1{10000}.$$

Answer 2

$$-f(-t) \leq f(x+t)-f(x) \leq f(t)$$ for all $x,t$. Hence $$|f(x+t)-f(x)| \leq \max \{|f(t)|, |f(-t)|\}.$$ Putting $t=y-x$ this gives $$|f(y)-f(x)| \leq \max \{|f(y-x)|, |f(x-y)|.$$ There exists $\delta >0$ such that $|t| <\delta$ implies $|f(t)|=|f(t)-f(0)| <\frac 1 {10000}$. Hence $|f(y)-f(x)| <\frac 1 {10000}$ whenever $|x-y| <\delta$.