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If $f$ is continuous on $\mathbb{R}$, $f'(0)=1$ and $f(x+y)=f(x)f(y)$ for all $x \in\mathbb{R}$, show that $f'(x)=f(x)$ for all $x\in\mathbb{R}$.

Solution: It is clear that $f(0)=1$. For each $x$ we have: \begin{eqnarray} \lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\cdots=f(x) \end{eqnarray}Hence, $f'(x)=f(x)$

My question: Where we use the continuity? Is it not enough that the function is defined on $\mathbb{R}$?

Jose Avilez
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Mathmath
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  • You are only provided that $f$ is continuous, you cannot assume the limit exists. – xbh Aug 19 '18 at 17:58
  • the limit exists becuase $f'(0)$ exists – Mathmath Aug 19 '18 at 17:59
  • Hint: use the conditions provided to explicitly find what $f(x) $ is. – xbh Aug 19 '18 at 18:00
  • No, the limit does not need exist for point $x$. For $f'(0)$, the limit should be $$ \lim_{h\to 0} \frac {f(h)-f(0)}h. $$ – xbh Aug 19 '18 at 18:01
  • Notee that $x$ is fixed – Mathmath Aug 19 '18 at 18:02
  • I do not see that, could you tell me how you discovered it? – xbh Aug 19 '18 at 18:03
  • No, $x$ is not fixed. The full statement of your question needs some universal quantifiers sprinkled in appropriately. For example, the hypothesis should say that $f(x+y)=f(x)f(y)$ for all $x,y \in \mathbb R$. Also, the conclusion should say that $f'(x)=f(x)$ for all $x \in \mathbb R$. – Lee Mosher Aug 19 '18 at 18:04
  • @Lee Mosher I corrected it, thank you – Mathmath Aug 19 '18 at 18:08
  • Well, if $f$ is continuous at a (any) single point and $f(x) f(y) =f(x+y) $ then $f$ is differentiable everywhere and $f'(x) =f'(0)f(x)$. So one can say that the existence of $f'$ at $0$ is redundant if one assumes continuity. See https://math.stackexchange.com/a/1885860/72031 – Paramanand Singh Aug 19 '18 at 20:11

3 Answers3

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By hypothesis

$$ \lim_{h\to0 }\frac{f(h)-f(0)}{h}=f'(0)=1. $$

So, we use that $f$ is differentiable at $0.$ And we don't need more assumptions about continuity or differentiability.

First of all note that $f(0)=1.$ $f(0+0)=f(0)^2\implies f(0)=1$ or $f(0)=0.$ If $f(0)=0$ then $f(x+0)=f(x)f(0)=0$ which contradicts the hypothesis $f'(0)=1.$

Now, using algebraic operations and that $f$ is differentiable at $0,$ we get $$ f'(x)=\lim_{h\to0 }\frac{f(x+h)-f(x)}{h}=\lim_{h\to0 }\frac{f(x)f(h)-f(x)f(0)}{h}=f(x)\lim_{h\to0 }\frac{f(h)-f(0)}{h}. $$

mfl
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Sorry, maybe you were right. I was assertive. If the continuity of $f$ is omitted, then for each $x$, $f(x+h)- f(x) = f(x)(f(h)- f(0)) \to 0 \cdot f(x) = 0 [h \to 0]$ since $f'(0)$ exists then $f$ is continuous at $0$, thus $f$ is still continuous everywhere.

Conclusion: $f \in \mathcal C (\mathbb R)$ is redundant.

xbh
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You don't need continuity for that. Since $f'(0)=1$, then, for each $x\in\mathbb R$,\begin{align}f'(x)&=\lim_{x\to0}\frac{f(x+h)-f(x)}h\\&=\lim_{x\to0}\frac{f(x)f(h)-f(x)}h\\&=f(x)\lim_{h\to0}\frac{f(h)-1}h\\&=f(x)\lim_{h\to0}\frac{f(h)-f(0)}h\\&=f(x)f'(0)\\&=f(x).\end{align}Note that $f(0)=f(0+0)=f^2(0)$ and that therefore $f(0)=0$ or $f(0)=1$. But you can't have $f(0)=0$, because otherwise$$(\forall x\in\mathbb{R}):f(x)=f(x+0)=f(x)f(0)=0$$and it would be false that $f'(0)=1$.

José Carlos Santos
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