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I need to prove that if $f(a+b)=f(a)f(b)$ for any two real numbers $a$, $b$, and $f(0)=f'(0)=1$, then $f(x)=f'(x)$ for all real number $x$.

So here's what I tried but it didn't quite work and I can't find a way to get the desired result;

Let $x,c$ be any real number,

$f(x)=f((\frac{x}{2}+c)+(\frac{x}{2}-c))=f(\frac{x}{2}+c)f(\frac{x}{2}-c)$

$f'(x)=\frac{1}{2}f(\frac{x}{2}+c)f'(\frac{x}{2}-c) + \frac{1}{2}f(\frac{x}{2}-c)f'(\frac{x}{2}+c)$

Substitute $f'(0)=1$, we get $1=\frac{1}{2}f(c)f'(-c) + \frac{1}{2}f(-c)f'(c)$

Hehe New User
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  • Hint: $f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$. – Daniel Schepler May 06 '17 at 15:03
  • this is the Cauchy equation, try google to solve it – Dr. Sonnhard Graubner May 06 '17 at 15:07
  • Just using the continuity of the function $f$ at any single point $a$ it is possible to prove that $f$ is continuous everywhere and differentiable everywhere too. And the behavior of function depends specifically on the value of its derivative $f'(0)$. See related https://math.stackexchange.com/a/1885860/72031 – Paramanand Singh May 07 '17 at 06:59

2 Answers2

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Fix a point $a$. For all $h\ne0$, $$f(a+h)-f(a)=f(a)(f(h)-1)$$ Divide by $h$, and use $f$ differentiable at $0$ : $$\frac{f(a+h)-f(a)}{h}=f(a)\frac{f(h)-f(0)}{h}\xrightarrow[h\to0]{}f(a)\times 1$$ Which proves that $f'(a)$ exists and $f'(a)=f(a)$.

Nicolas FRANCOIS
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  • As far as I can see, this proves only that $f'(x) = f(x) f'(0)$. It was assumed in the proof that $f'(0)=1$, which is not true for the function $g(x) = e^{ax}$. This function, obviously, fulfills the assumption that $g(x+y) = g(x) g(y)$, but $g'(x) = a \neq 1$. Thus, the statement $f(x+y) = f(x) f(y) \Longrightarrow f(x) = e^x$ looks incorrect. It should be $f(x+y) = f(x) f(y) \Longrightarrow f(x) = e^{ax}$, with $a \in \mathbb{R}$. – QuantumBrick May 03 '22 at 12:09
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    In the OP, the assumption is $f(0)=f'(0)=1$... – Nicolas FRANCOIS May 04 '22 at 14:49
  • True. I didn't see that! – QuantumBrick May 05 '22 at 15:11
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You can't use $f'(x)$ before having proved it exists.

Consider instead that $$ f(x+h)-f(x)=f(x)f(h)-f(x)= f(x)\bigl(f(h)-1\bigr)=f(x)\bigl(f(0+h)-f(0)\bigr) $$

egreg
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