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How can I prove that if a function is such that $f'(x)=f(x)$ and also $f(0)=0$ then $f(x)=0$ for every $x$. I have an idea but it's too long, I want to know if there is a simple way to do it. Thanks! Obviously in a formal way.

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    $\begingroup$ What about that one? $\endgroup$ – Ilya May 30 '13 at 12:51
  • $\begingroup$ I believe this is an exercise in Bartle. $\endgroup$ – gt6989b May 30 '13 at 13:36
  • $\begingroup$ Why don't you explain the idea you have? $\endgroup$ – Did May 30 '13 at 15:56
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An implicit assumption is that the function is defined on some open interval containing $0$. Set

$$g(x)=e^{-x}f(x)$$

and compute the derivative:

$$g'(x)=-e^{-x}f(x)+e^{-x}f'(x)=-e^{-x}f(x)+e^{-x}f(x)=0$$

so the function $g$ is constant on the interval where it's defined. Since $g(0)=e^{-0}f(0)=0$ you can conclude that $g(x)=0$ for all $x$ and therefore also $f(x)=0$ for all $x$.


Without the initial assumption, you can get different functions with that property: define, for instance, $$f(x)=\begin{cases} 0 & \text{if $-1<x<1$}\\ e^x & \text{if $2<x<3$} \end{cases}$$ Then $f$ satisfies the requirements, but it's not constant.

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    $\begingroup$ Here $f(x)$ is not continuous so it is not differentiable hence no question of $f'=f$... $\endgroup$ – Abhra Abir Kundu May 30 '13 at 13:06
  • $\begingroup$ @AbhraAbirKundu Are you really sure that the function $f$ in the counterexample is not continuous? I am sure it is. Please, state where the function is not continuous. $\endgroup$ – egreg May 30 '13 at 13:07
  • $\begingroup$ @egreg I guess it's a matter of casuistics, $f$ is left-continuous at $x=1$ but not right-continuous, since it does not exist. Usually with standard definitions, we define continuiuty to be 2-sided. $\endgroup$ – gt6989b May 30 '13 at 13:26
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    $\begingroup$ I see. Still something tells me OP was looking for the functions defined on the entire real line, at least that's the way I read "for every $x$" - but I see how that could be interpreted differently. $\endgroup$ – gt6989b May 30 '13 at 13:30
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    $\begingroup$ @gt6989b I agree about OP's probable intentions. But the problem should be stated differently, saying explicitly that the function is defined on the whole real line. The final "for all $x$" means "for all $x$ where $f$ is defined" and the stated assumption merely require that $f$ is defined in an open neighborhood of $0$ (to ensure differentiability in the domain). $\endgroup$ – egreg May 30 '13 at 13:33
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We can solve differential equation $f'(x)-f(x)=0$. It's linear differential equation with constant coefficient, and we know how to solve it. Setting $f(x)=e^{\lambda x}$ we obtain $\lambda e^{\lambda x} - e^{\lambda x}=0$, or $\lambda - 1 =0$, which implies $\lambda = 1$. So our solution is $f(x)=c e^x$. But, we have condition $f(0)=0$, so $c e^0=0 \implies c=0$, so our final solution is $f(x)=0$.

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    $\begingroup$ More direct version: let $g(x)=e^{-x}f(x)$. Then $g(0)=0$ and $g'(x)=-e^{-x}f(x)+e^{-x}f'(x)=0$, so $g=0$ and $f=0$. $\endgroup$ – wj32 May 30 '13 at 13:00
  • $\begingroup$ @wj32 We had the same idea. $\endgroup$ – egreg May 30 '13 at 13:04
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Here is a different approach, more suited for calculus students who do not know how to solve differential equations yet.

It's easy to prove by induction that since $f'=f$, the function $f$ must have derivatives of all orders and $f = f' = f'' = ...$ for all $x$, and hence $f(0) = f'(0) = f''(0) = \ldots$.

Then, by Taylor's Theorem, we expand $f(x)$ in series around 0 to get

$$ f(x) = \sum_{n=0}^\infty \frac{x^n}{n!} f^{(n)}(0) = \sum_{n=0}^\infty \frac{x^n}{n!} \cdot 0 = 0. $$

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    $\begingroup$ How do you know that $f$ is expandable as its Taylor series? $\endgroup$ – egreg May 30 '13 at 13:37
  • $\begingroup$ @egreg Since it is infinitely differentiable at every point. $\endgroup$ – gt6989b May 30 '13 at 13:38
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    $\begingroup$ There exist functions that are infinitely diffentiable at every point, but aren't the sum of their Taylor series at every point; the classical example is $f(x)=e^{-1/x^2}$ for $x\ne0$ and $f(0)=0$. $\endgroup$ – egreg May 30 '13 at 13:43
  • $\begingroup$ @egreg You are right. I remember the problem from years ago as an exercise in Bartle Elements of Real Analysis - he did spend a serious effort trying to get rid of this problem, but I don't remember how. $\endgroup$ – gt6989b May 30 '13 at 13:53
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Below I show how the standard proof (e.g. egreg's answer) is a special case of more general results on uniqueness theorems and Wronskians. Let's rewrite the proof in slightly more general form. Given that $\,f\,$ and $\,g = e^x\,$ are solutions of $\,y' = y\,$ we deduce

$$\begin{eqnarray}\color{#c00}{f'=f}\\ \color{#0a0}{g'=g}\end{eqnarray}\ \ \Rightarrow\ \ \left(\dfrac{f}g\right)' = \dfrac{\color{#c00}{f'}g-f\color{#0a0}{g'}}{g^2} = \dfrac{\color{#c00}fg-f\color{#0a0}g}{g^2} = 0\ \ \Rightarrow\ \dfrac{f}g = c\ \ \rm constant$$

Therefore $\ f = cg = ce^x\ $ so $\ 0 = f(0) = c,\ $ thus $\, f = 0.\, $ This generalizes. The same proof shows that, assuming appropriate differentiability/continuity conditions, if $\,f,g\,$ are solutions of $\, y' = h y\,$ on an interval $\,I\,$ where the Wronskian $\,W(f,g) := f'g - fg' = 0\,$ on $\,I,\,$ then $\,f,g\,$ are linearly dependent on $\,I,\,$ i.e. $\,c_1 f = c_2 g \,$ on $\,I\,$ for some constants $\,c_i.\,$

Thus, by a very simple proof, we've deduce a uniqueness theorem for solutions of linear first-order differential equations of said type. These ideas extend to analogous higher-order linear differential equations. See here for a proof of the second-order case (which generalizes to n'th order), using variation of parameters, and see here for the discrete analog of difference equations (recurrences). See also the classical result below on Wronskians and linear dependence.

Theorem $\ \ $ Suppose $\rm\:f_1,\ldots,f_n\:$ are $\rm\:n-1\:$ times differentiable on interval $\rm\:I\subset \mathbb R\:$ and suppose they have Wronskian $\rm\: W(f_1,\ldots,f_n)\:$ vanishing at all points in $\rm\:I\:.\:$ Then $\rm\:f_1,\ldots,f_n\:$ are linearly dependent on some subinterval of $\rm\:I\:.$

Proof $\ $ See this answer.

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  • $\begingroup$ Seriously? If you want to hit it with differential equations why don't you use local lipchitz continuous? That would be 100 times easier? $\endgroup$ – Dominic Michaelis May 30 '13 at 18:16
  • $\begingroup$ @Dominic I edited the answer to make it clear that this simple proof requires no advanced knowledge - it can be understood by any calculus student. I mention more general uniqueness theorems for differential equations etc only to emphasize that this is a special case of more general results. I don't use these more general results in my proof. An important part of teaching is highlighting the relationships between concrete and general results. $\endgroup$ – Key Ideas May 30 '13 at 19:55
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Assuming $f$ is continuous and infinitely differentiable:

  • If $f(x) = f'(x)$, then: $f'(x) = f''(x)$, $f''(x) = f'''(x)$, et cetera.

  • One function satisfies this property... Or rather, is defined by this property: $f(x) = A\,e^x$.

  • $A$ can be determined by evaluating $f$ at $x=0$ such that $A = f(0)$.

  • If $f(0) = 0$, then $A = 0$, therefore $f(x) = 0$ for any $x$.

In summary:

  1. $f(x) = f^{\prime}(x) \rightarrow f(x) = A\,e^x$, or more generally $\lambda f(x) = f^{\prime}(x) \rightarrow f(x) = A\,e^{\,\lambda x}$

  2. $f(0) = 0 \rightarrow A = 0$, or more generally, $f(k) = 0 \rightarrow A = 0$

  3. $A = 0 \rightarrow f(x) = 0$

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