Show that $~f~$ is uniformly continuous on all of $~\mathbb R~$.

Question

Suppose $~f: \mathbb R\to \mathbb R~$ is a function that for every $~x,~y~$ in $~\mathbb R~$, $f(x+y)=f(x)+f(y)~$ and $~f~$ is continuous at $~0~$. Show that $~f~$ is uniformly continuous on all of $~\mathbb R~$.

I tried proving that for every epsilon > 0, there exists delta > 0 such that |x-x0| < delta and |f(x)-f(x0)| < epsilon but I did not know really what to do with the given information.

Answer 1

Note the following

• $f(0) = f(0+0) = 2f(0) \Rightarrow f(0) = 0$
• $\lim_{h\to 0}f(h) = f(0) = 0 \Rightarrow \forall \epsilon > 0 \exists \delta >0: |h|<\delta \Rightarrow |f(h)| < \epsilon$

Now, for any $x\in \mathbb{R}$ you have: $$|f(x+h)-f(x)| \stackrel{f(x+h) = f(x) + f(h)}{=} |f(h)| < \epsilon \mbox{ for } |h| < \delta$$

Since $\epsilon$ is independent of $x$, this is uniform continuity.

Answer 2

Alternatively, you $\it{could}$ attempt to "solve: for $f$:

Given $f(x+y) = f(x)+f(y)$ and $f$ continuous at $0$, you can then try to see which known functions satisfy the given equation. Such an equation is called a functional equation.

To find out what $f$ is at $0$, we do as in the answer given by trancelocation and get that:

$f(0)=f(0+0)=2f(0) \implies f(0)=0$

Next, we solve for $f$. A good educated guess will suggest that $f(x)=cx$, i.e. a linear equation. This is indeed correct.

Finally, you can argue (similar as done in trancelocation's answer) that $f(x)=cx$ is uniformly continuous and you are done.

For more on functional equations, see:

https://brilliant.org/wiki/functional-equations/ https://artofproblemsolving.com/wiki/index.php/Cauchy_Functional_Equation

Hopefully this is interesting and adds a new perspective :)

(Note: I used the assumption that $f$ is continuous at $0$ to get the function as $f(x)=cx$. Without this assumption, there can also be other functions which are (heavily) discontinuous.)