User:J D Bowen/For Julia

$$f(x+y) = f(x)f(y) \ $$

First note that $$f(1)=f(0+1)=f(0)f(1) \implies f(0)=1 \ $$.

We want to show $$f \ $$ continuous at zero implies $$f \ $$ is cont at c.

That means we need to solve $$|x-c|\leq\delta \implies |f(x)-f(c)|\leq \epsilon \ $$ for $$\delta \ $$ in terms of $$\epsilon \ $$.

So, at worst $$x=c\pm \delta \ $$.

Then $$|f(c)-f(x)|=| f(c)-f(c)f(\pm\delta)| = |f(c)(1- f(\pm\delta))| \leq \epsilon $$.

This will be true if, given $$\epsilon \ $$, we can always find a $$\delta \ $$ such that $$|1 - f(\pm\delta)|\leq \frac{\epsilon}{|f(c)|} \ $$. But since $$1=f(0) \ $$, that's the same as requiring that we can always find a delta such that $$|f(0)-f(\pm\delta)|\leq \frac{\epsilon}{|f(c)|} \ $$. But this is precisely what it means to be continuous at $$0 \ $$.