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Does a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f'(x) > f(x) > 0$ exist?

Intuitively, I think it can't exist.

I've tried finding the answer using the definition of derivative:

  1. I know that if $\lim_{x \rightarrow k} f(x)$ exists and is finite, then $\lim_{x \rightarrow k} f(x) = \lim_{x \rightarrow k^+} f(x) = \lim_{x \rightarrow k^-} f(x)$

  2. Thanks to this property, I can write:

$$\begin{align} & f'(x) > f(x) > 0 \\ & \lim_{h \rightarrow 0^+} \frac{f(x + h) - f(x)}h > f(x) > 0 \\ & \lim_{h \rightarrow 0^+} f(x + h) - f(x) > h f(x) > 0 \\ & \lim_{h \rightarrow 0^+} f(x + h) > (h + 1) f(x) > f(x) \\ & \lim_{h \rightarrow 0^+} \frac{f(x + h)}{f(x)} > h + 1 > 1 \end{align}$$

  1. This leads to the result $1 > 1 > 1$ (or $0 > 0 > 0$ if you stop earlier), which is false.

However I guess I made serious mistakes with my proof. I think I've used limits the wrong way. What do you think?

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    $\begingroup$ $f(x)=e^{2x}$.${}$ $\endgroup$
    – David Mitra
    Feb 2, 2015 at 22:20
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    $\begingroup$ @DavidMitra why not make that an answer? $\endgroup$
    – Steven Gubkin
    Feb 2, 2015 at 22:21
  • $\begingroup$ Do you want your condition to hold over all of $\mathbb{R}$? There is certainly a counterexample over $(1, \infty)... $\endgroup$
    – Steven Stadnicki
    Feb 2, 2015 at 22:21
  • $\begingroup$ Comments can be used when the question should be edited (or in this case extremely edited by removing it). Since this is the answer that was in my mind after about two seconds, David possibly didn't think this question needed an answer. $\endgroup$
    – gnasher729
    Feb 3, 2015 at 18:14

6 Answers 6

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expanded from David's comment

$f' > f$ means $f'/f > 1$ so $(\log f)' > 1$. Why not take $\log f > x$, say $\log f = 2x$, or $f = e^{2x}$.

Thus $f' > f > 0$ since $2e^{2x} > e^{2x} > 0$.

added: Is there a sub-exponential solution?
From $(\log f)'>1$ we get $$ \log(f(x))-\log(f(0)) > \int_0^x\;1\;dt = x $$ so $$ \frac{f(x)}{f(0)} > e^x $$ and thus $$ f(x) > C e^x $$ for some constant $C$ ... it is not sub-exponential.

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    $\begingroup$ Great answer - wondering if there is a "sub-exponential" answer. $\endgroup$
    – Ryan Dougherty
    Feb 3, 2015 at 18:53
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There are multiple mistakes in your proof (i.e. dividing by $f(x)$ is not necessarily okay, since it is not necessarily positive). The most major is that you treat the variable in the limit as if it were not "bound". That is, if you have something like $$\lim_{h\rightarrow 0^+}1>0$$ which is true, you can't necessarily, say, multiply through by $h$ to get $$\lim_{h\rightarrow 0^+}h>0\cdot h$$ which is false. This is essentially what you do, and why your conclusion is wrong. You have to regard the $h$ as belonging to the limit - that is $\lim_{h\rightarrow 0^{+}}f(h)$ is a constant - it does not depend on $h$, because there is no notion of "$h$" outside of the limit.

For an example of a function that does not satisfy this, you can take $e^{\alpha x}$ for any $\alpha>1$. Another solution is $xe^x$. It's worth noting that since the solution to $f'=f$ is $x\mapsto e^x$ we can prove that any solution grows faster than exponentially.

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    $\begingroup$ The important thing to keep in mind is that relational symbols like $>$, $<$ and $=$ have lower precedence than $\lim$. That is $\lim_{h\to0^+} h > 0\cdot h$ may be equivalently written as $\left( \lim_{h\to0^+} h \right) > (0\cdot h)$ -- and now it's obvious that this expression is nonsense, since the $h$ on the right-hand side is outside the scope of the $\lim$ on the left-hand side, and is thus an undefined free variable. $\endgroup$
    – Ilmari Karonen
    Feb 2, 2015 at 23:23
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    $\begingroup$ ... This is actually IME a fairly common notational confusion among students starting to learn limits: some will mistakenly parse the limit as somehow applying to the whole (in)equation, as if it were written $\lim_{h\to0^+}(h > 0 \cdot h)$, and try to interpret it as something like "in the limit as $h$ tends to $0$ from above, $h > 0 \cdot h$" (which, of course, is technically a true statement, if we're liberal enough with our notation to allow taking limits of truth-valued expressions like $(x>y)$, but also a pretty useless one). $\endgroup$
    – Ilmari Karonen
    Feb 2, 2015 at 23:31
  • $\begingroup$ In the terminology I use, for $\alpha>1$, the function $x \mapsto e^{\alpha x}$ grows exponentially, not "faster than exponentially". Of course its quotient by $e^x$ still diverges ... exponentially. $\endgroup$
    – Jeppe Stig Nielsen
    Feb 3, 2015 at 0:49
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    $\begingroup$ I will point out that if for every neighborhood $N$ of $x_0$, there exists a $y \in N$ such that $f(y) > g(y)$, then $\lim_{x\to x_0} f(x) \ge \lim_{x\to x_0} g(x)$, provided both limits exist. Thus, you can redeem your argument by replacing the second $>$ with $\ge$. Done in the OP's work, this becomes $1 \ge 1$, which removes the apparent contradiction. $\endgroup$
    – user88319
    Feb 3, 2015 at 1:22
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Even more generally than what other answers have brought up (though this implicit in a step of GEdgar's proof), if you let $g$ be a continous function, and $G$ an arbitary primitive of $g$ (that is, G' = g), then if $f(x) = e^{G(x)}$, we have $f' = gf$. So the ratio of a function to it's derivative can grow at essentially arbitary rates.

Edit: as Marc van Leeuwen points out, an example gotten from this is that $f(x) = e^{x^2}$ satisfies the property that $f'(x)/f(x) = 2x$ so the derivative grows asymptotically faster than the function.

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    $\begingroup$ You might add that hence an obvious function whose derivative grows asymptotically faster than itself is $x\mapsto\exp(x^2)$. $\endgroup$
    – Marc van Leeuwen
    Feb 3, 2015 at 8:45
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Let's keep working with your reasoning. If $f'(x) > f(x)$, then we have that for $h > 0$ sufficiently small,

$$\frac{f(x+h) - f(x)}{h} > f(x)$$

so

$$f(x+h) - f(x) > hf(x)$$ $$\frac{f(x+h)}{f(x)} > h+1$$

As you found. The problem occurs when you take the limit. The correct way to take the limit is

$$\lim_{h \to 0^+} \frac{f(x+h)}{f(x)} \ge \lim_{h \to 0^+} h+1$$ $$1 \ge 1$$

which is not a contradiction. To see why we have to take the limit like this, let's look at the functions $f(x) = x$, $g(x) = 1$. Now, we can certainly argue that for $x < 1$, $f(x) < g(x)$; however,

$$\lim_{x\to1^-}f(x) = 1 = \lim_{x\to1^-}g(x)$$

As others have already pointed out, what you are trying to prove is false: $f(x) = e^{2x}$ is a counterexample.

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Let $f(x) =e^x + C $ where $C<0$ then $f'(x)-f(x)=-C>0$ or $f'(x)>f(x)$ for all x - a trivial example. In this example $f(x)$ can be uniformly >, or < $f(x)$ depending on thee value of $C$.

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  • $\begingroup$ Elegant but the OP asks for a solution that works for $\mathbb{R}$ and is $>0$... $\endgroup$
    – Martigan
    Feb 4, 2016 at 17:21
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My personal favorite counterexample is $x^x$. When you differentiate it, you get $x^x[ln(x)+1]$, which is greater than $x^x$ over $[0,\infty]$ (which is the largest interval over which the derivative is defined). And, while it's not possible to get a closed form integral, doing some numerical integration, maybe in Mathematica, shows that the integral grows more slowly than the function itself. Of course, we would have to do quite a bit more work to actually prove this.

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