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,


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

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

  • $\begingroup$ Hint: $f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$. $\endgroup$ – Daniel Schepler May 6 '17 at 15:03
  • $\begingroup$ this is the Cauchy equation, try google to solve it $\endgroup$ – Dr. Sonnhard Graubner May 6 '17 at 15:07
  • $\begingroup$ 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 math.stackexchange.com/a/1885860/72031 $\endgroup$ – Paramanand Singh May 7 '17 at 6:59

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 $f'(a)=f(a)$.


You can't use $f'(x)$ before having proved it exists.

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


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