I am currently studying for a qualifying exam, and I came across the following question:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(x+y) = f(x)+f(y)$ for all $x,y$. Suppose that there exists $a,M>0$ such that $|f(x)|\leq M$ for all $x \in [-a,a]$. Prove that $f$ is uniformly continuous.
I am not sure how to approach this question. I have attempted to use an epsilon-delta argument, but that did not prove effective. I also thought that if we could show that $f$ is continuous on $[-a,a]$, then we would have uniform continuity by the compactness of $[-a,a]$. However, this still wouldn't give us that $f$ is uniformly continuous everywhere. Any suggestions would be greatly appreciated.
Claim #1: $f(0)=0$ and $f$ is continuous at $x=0$
Proof of Claim #1: Notice $f(0)=f(0+0)=f(0)+f(0)=2f(0)$ which means $f(0)=0$. Now let $\epsilon>0$ be arbitrary. Find a positive integer $n$ so that $\frac{M}{n}<\epsilon$, then put $\delta:=\frac{a}{n}$. Notice $x\in (-\delta,\delta)$ implies $nx\in(-a,a)$ and so $$|f(x)|=\Bigg|\frac{f(x)+\dots+f(x)}{n}\Bigg|=\Bigg|\frac{f(nx)}{n}\Bigg|\leq \frac{M}{n}<\epsilon$$ This shows $f$ is continuous at $x=0$.
Claim #2: $f(-x)=-f(x)$ for all $x\in \mathbb{R}$
Proof of Claim #2: We get $f(x)+f(-x)=f\big(x+(-x)\big)=f(0)=0$ so $f(-x)=-f(x)$.
Claim #3: $f$ is uniformly continuous on $\mathbb{R}$.
Proof of Claim #3: Since $f$ is continuous at $0$, we can find $\delta>0$ so that $|x|<\delta$ implies $|f(x)|<\epsilon$. Choose $x,y\in \mathbb{R}$ such that $|x-y|<\delta$. Then $$|f(x)-f(y)|=|f(x)+f(-y)|=|f(x-y)|<\epsilon$$ and we're done.