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 $$0 \ $$.

Let $$f(1)=a \ $$.

Observe that for $$n\in\mathbb{N} \ $$, we have $$f(n)=f(\underbrace{1+\dots+1}_{n \ \text{times}}) = \underbrace{f(1)...f(1)}_{n \ \text{times}}=a^n \ $$.

Further observe that $$f(1/n)^n = \underbrace{f(1/n)...f(1/n)}_{n \ \text{times}} = f(\underbrace{1/n+\dots+1/n}_{n \ \text{times}})=f(1)=a \ $$, and so $$f(1/n)=a^{1/n} \ $$.

Notice that for any positive rational $$m/n \ $$, we have $$f(m/n)=f(\underbrace{1/n+\dots+1/n}_{m \ \text{times}}) = \underbrace{f(1/n)...f(1/n)}_{m \ \text{times}}=a^{m/n} \ $$.

Moving to the negatives, note that $$1=f(0)=f(1-1)=f(1)f(-1)=af(-1) \implies f(-1)=a^{-1} \ $$.

Observe that for positive $$n \ $$, we have $$f(-n)=f(-1-1+\dots-1) = f(-1)^n = a^{-n} \ $$.

We also note $$f(-1/n)^n = f(-1/n)f(-1/n)\dots f(-1/n) = f(-1/n - \dots -1/n) = f(-1)=a^{-1} \implies f(-1/n)=a^{-1/n} \ $$.

Finally, note for any positive rational $$m/n \ $$, we have $$f(-m/n)=f(\underbrace{-1/n-\dots-1/n}_{m \ \text{times}}) = \underbrace{f(-1/n)...f(-1/n)}_{m \ \text{times}}=a^{-m/n} \ $$.