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Hi guys I wondered whether you could help me to prove the following, This is part of a longer exam question which I'm revising now. Also could you recommend any good books with proofs relating to calculus please.

Suppose $f:\mathbb R \to \mathbb R$ is a continuous function such that:

  • $f$ is differentiable at $0$ with $f(0) = 1$ and $f'(0)=1$
  • $f(s+t) = f(s)f(t) $ for all $s, t\in \mathbb R$

Prove that $f(x)>0$ for all $x\in \mathbb R$. Prove that $f$ is differentiable on $\mathbb R$ with $f'(x)=f(x)$ for $x\in \mathbb R$. Deduce that $f(x)e^{-x}$ is constant, and hence $f(x)=e^x$

Thank-you :)

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  • $\begingroup$ For future reference: you can get the $\in$ symbol using \in $\endgroup$ – Winther Dec 31 '16 at 18:41
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For the first part, set $t=-s$ to deduce that $1=f(0)=f(s)f(-s)$ for all $s\in\mathbb{R}$. Therefore $f$ is never zero. By continuity, it is either always positive or always negative, but $f(0)>0$, so it's always positive.

For the second part, notice that $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{f(x)f(h)-f(x)}{h}=\lim_{h\to0}\frac{f(x)\left(f(h)-1\right)}{h}=f(x)\color{blue}{\lim_{h\to0}\frac{f(h)-1}{h}}$$

Do you recognize the limit in blue?

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  • $\begingroup$ (I was hoping to leave the rest of the proof to you, Curtis, but Daminark gave it away.) $\endgroup$ – symplectomorphic Dec 31 '16 at 18:54
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We know that $f(x) = f(\frac{x}{2} + \frac{x}{2}) = f(\frac{x}{2})^2$. Now, if $f(x) = 0$ for some $x\in \mathbb{R}$, then $f(0) = f(x-x) = f(x)f(-x) = 0 \ne 1$, so that $f(x) > 0$ for all $x\in \mathbb{R}$. Now, taking the definition of the derivative: $$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = f(x)\lim_{h\to 0} \frac{f(h)-1}{h} = f(x)f'(0) = f(x)$$ Knowing this, we see that the function $g(x) = f(x)e^{-x}$ has derivative $0$, by product rule, so it is constant. Therefore, $f(x) = ce^x$ for all $x$. Using the values for $x=0$, we see that $c=1$.

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  • $\begingroup$ Where did you use the fact that $\forall x\in \mathbb R\left(f(x)=\left(f\left(\frac x 2\right)\right)^2\right)$? $\endgroup$ – Git Gud Dec 31 '16 at 18:46
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    $\begingroup$ That gives you that $f(x)$ is at least 0, so coupled with how $f(x)$ can't be $0$ anywhere gives us that it's strictly positive. $\endgroup$ – Daminark Dec 31 '16 at 18:48
  • $\begingroup$ Ah, I see. Thanks. $\endgroup$ – Git Gud Dec 31 '16 at 18:49
  • $\begingroup$ Yeah, nice: you give an algebraic proof that $f$ is positive, while I give a topological one. $\endgroup$ – symplectomorphic Dec 31 '16 at 18:50
  • $\begingroup$ Oh huh, I never would've thought of proving it like that. $\endgroup$ – Daminark Dec 31 '16 at 18:53
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If $f:\mathbb{R} \to \mathbb{R}$ is continuous at a single point say $c$ and satisfies $$f(s+t) =f(s) f(t) $$ for all real $s, t$ then it is possible to prove that $f$ is differentiable everywhere with $f'(x) =f(x) f'(0)$. Moreover we can also prove that either $f(x) =0$ for all values of $x$ or $f(x) =a^{x} $ for some specific positive number $a$ namely $a=f(1)$. The value of $a$ is linked with the value of $f'(0)$ and in fact we can show that $f'(0)=\log a$ so that if $f'(0)=1$ then $a=e$. See this answer https://math.stackexchange.com/a/1885860/72031 for more details.

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