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Let $f:\mathbb R \to \mathbb R$ be a function which is differentiable at every $x \in \mathbb R$.

a) Assume that $f'(x) = f(x)$ for all $x \in \mathbb R$ and $f(0) = 0$. Prove that $f(x) = 0$ for all $x \in \mathbb R$. (Hint: Taylor's theorem.)

b) Assume that $f'(x) = f(x)$ for all $x \in \mathbb R$. Prove that $f(x) = f(0)e^x$.

Kenny Wong
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mathdude
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  • $ f'(x)=f(x) \implies f^{(n)}(0)=f(0) $ – WW1 May 02 '17 at 02:27
  • @WW1: just because all derivatives of $f$ are zero at zero does not mean that $f$ is the zero function. The classic example is $e^{(x^{-2})}$. It has all derivatives zero at $x=0$ but is not constant. – Ross Millikan May 02 '17 at 02:41
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    @RossMillikan I believe you mean $e^{\left(-x^{-2}\right)}$ – MathematicsStudent1122 May 02 '17 at 02:56
  • for that function - it is true that $$ \lim_{x\to 0}f^{(n)}(x)=0$$ but $f^{(n)}(x)$ has a removable discontinuity at $x=0$ so $f$ is not technically differentiable at $x=0$ – WW1 May 02 '17 at 03:12
  • @WW1: we can remove it, as you say. Then it is infinitely differentiable and all derivatives are zero. – Ross Millikan May 02 '17 at 03:34
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    @MathematicsStudent1122: you are right. I meant $e^{-x^{-2}}$, with f(0)=0$ – Ross Millikan May 02 '17 at 03:35
  • @RossMillikan that would be true if only $f(x)$ has a removable discontinuity, but in this case $f^{(n)} (x)$ itself has a removable discontinuity. $\$ Wolframe Alpha agrees ... https://www.wolframalpha.com/input/?i=f(x)%3De%5E(-x%5E-2)+what+is+f%27(0) – WW1 May 03 '17 at 02:47
  • see this answer https://math.stackexchange.com/a/1292586/72031 We don't need any powerful theorems like Taylor. Mean value theorem is sufficient to handle the problem. – Paramanand Singh May 03 '17 at 06:44

1 Answers1

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a) By the assumption $f'(x)=f(x)$, we can conclude that $f$ is infinitely differentiable with $$f^{(n)}(x)=f(x),\quad\forall x\in\mathbb{R}, n\in\mathbb{N}.$$ For $x>0$, and $k\in\mathbb{N}$, by Taylor's theorem there exists $\xi_k\in(0,x)$ such that $$f(x)=\sum_{j=0}^k\frac{f^{(j)}(0)}{j!}x^j+\frac{f^{(k+1)}(\xi_k)}{(k+1)!}x^{k+1}=\frac{f(\xi_k)}{(k+1)!}x^{k+1}.$$ Set $M_x=\max_{t\in[0,x]}|f(t)|$, we have $$|f(x)|\leq \frac{M_x}{(k+1)!}x^{k+1}\to 0,\text{ as }k\to\infty,$$ Therefore $f(x)=0$. Similarly we have $f(x)=0$ when $x<0$. Therefore $f(x)=0$ for all $x\in\mathbb{R}$.

b) From the proof of a), one can see that if $f$ vanishes at some point, then $f$ must be identically $0$ on $\mathbb{R}$, and in this case clearly $$0=f(x)=f(0)e^x.$$ Now assume $f$ never vanishes, then by $f'(x)=f(x)$ we have $$\frac{f'(x)}{f(x)}=1,$$ taking anti-derivative on both sides we have $$\log|f(x)|=x+C,$$ where $C$ is some constant. This means $$|f(x)|=e^Ce^x,\quad\Rightarrow\quad f(x)=\pm e^Ce^x.$$ Plug in $x=0$, we have $$f(0)=\pm e^C.$$ Hence $$f(x)=f(0)e^x.$$

An alternative way to do b): As Winther suggested, $f'(x)=f(x)$ is equivalent to $$f'(x)-f(x)=0.$$ Therefore we have $$0=e^{-x}f'(x)-e^{-x}f(x)=(e^{-x}f(x))'.$$ It follows that $$e^{-x}f(x)=C\quad \Rightarrow\quad f(x)=Ce^x$$ for some constant $C$. Take $x=0$ we get $C=f(0)$.

Frank Lu
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  • @Winther You're absolutely right, and actually this is better than my approach because there's no need to worry about where $f$ vanishes. Thanks for the nice suggestion. – Frank Lu May 02 '17 at 03:43
  • In this spirit of ODEs, I think a proof of unicity would go well in here. If $f_1,f_2:\mathbb{R}\longrightarrow\mathbb{R}$ are two functions that satisfy $f(0)=C$ and $f'=f$, then $g:=f_1-f_2$ satisfies $g'=g$ and $g(0)=0$. By the first exercise, $g\equiv 0$ throughout $\mathbb{R}$, so $f_1\equiv f_2$ throughout $\mathbb{R}$. In other words, for each $C\in\mathbb{R}$ there is a single solution to $f'=f$, $f(0)=C$. Finally, at this point it suffices to note that, for each $C$, $f(x)=Ce^x$ is a solution. – Fimpellizzeri May 02 '17 at 05:52