Continuity of a function

Question

For all $(x,y)\in\mathbb{R}^2$ a function $f$ satisfies $f (x+y)=f (x)\cdot f (y)$.

If $f$ is continuous at a point $a$, then show that $f$ is continuous on $\mathbb R$ and $f (x)= b^x$ for some constant $b$.

My approach

$$f (1)=f (1+0)=f (1)\cdot f (0)$$ Thus $f(0)=1$ and $$f(x+h)-f (x)= f(x)\,(f (h)-1)$$

I can prove the continuity of the function. How can I prove that $f (x)=b^x$ for any $x\in\Bbb R$?

See also the rest of the topic overview: https://math.stackexchange.com/q/423492/87023 — Chris Culter, Sep 03 '17 at 06:48

score 1 · Answer 1 · answered Sep 03 '17 at 06:32

1

Hint:

From the continuity of $f$ and $f(0)=1$, show that $f(1) > 0$. Then show that for any rational number $r$, we have $$f(r) = f(1)^r$$

answered Sep 03 '17 at 06:32

pisco

18,983

I can prove that for any natural number n f (n) =f (1)^n – Anil Kumar Pandey Sep 03 '17 at 06:39
But i cant get how to prove this any real number r – Anil Kumar Pandey Sep 03 '17 at 06:40
@AnilKumarPandey Note that $$f(\frac{n}{m}) = f(\frac{1}{m})^n $$ On the other hand, from the functional equation $$f(\frac{1}{m})^m = f(1)$$ So $f(r) = f(1)^r$ for any rational number $r$. This extends to all $r\in \mathbb{R}$ because $f$ is continuous. – pisco Sep 03 '17 at 06:42
thanks for valuable suggestions it really helped me – Anil Kumar Pandey Sep 03 '17 at 07:05

Continuity of a function

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