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Solve for $f: \mathbb{R}\to\mathbb{R}\ \ \ $ s.t.

$$f(n)=e^n \ \ \forall n\in\mathbb{N}$$ $$f^{(y)}(x)>0 \ \forall y\in\mathbb{N^*} \ \forall x\in\mathbb R$$

Could you please prove that there exists an unique solution: $f(x)=e^x$?

(Anyway, this problem is not about fractional calculus)

$\mathbb N^*=\{1,2,3...\}, \ \mathbb N=\{0,1,2....\}$

How about try to construct a few functional spaces that intersect at one point?

Try Sard Theorem and Pre image Theorem.

High GPA
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  • What do you mean by $f^{(y)}(x)>0~\forall y\in\mathbb R,y\ge1$? Particularly, when $y\notin\mathbb N$. – Simply Beautiful Art Jun 10 '17 at 12:27
  • Hm, it appears WA glides over the point that there are multiple definitions to the fractional derivative, many of which do not have to agree. – Simply Beautiful Art Jun 10 '17 at 12:31
  • So you want to prove that $e^x$ is the only function that has all of its derivative positive ? – Zubzub Jun 10 '17 at 13:22
  • I'm sorry, I misunderstood the question. I deleted my answer. – orlp Jun 10 '17 at 14:50
  • @Reveillark unfortunately not... $f''>0$ does not imply $(\ln f)'' >0$ – High GPA Jun 13 '17 at 14:13
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    What is $\Bbb N^*$? $\Bbb N$ with zero? – M. Winter Jun 13 '17 at 14:59
  • Woopsy daisy, you're right, that's what it get for just computing one derivative. – Reveillark Jun 13 '17 at 15:18
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    This is a pretty interesting problem.

    Here, $\mathbb{N}$ isn't that special. What I mean is that for any discrete and countably infinite set $S \subset \mathbb{R}$ one can ask a similar question but instead of $f(n) = e^n ,\ \forall n \in \mathbb{N}$, we have $f(c) = e^c \ ,\forall c \in S$. Indeed, it's possible the unique solution may hold even if $S$ is finite with $|S| \geq 2$, though I'm not sure of that.

     – MathematicsStudent1122 Jul 11 '17 at 03:34
  • In case it saves someone else some time: my first guess was to try to add a bump function supported on, say, $[2,3]$ to modify $e^x$ to get a counterexample. However, according to https://math.stackexchange.com/questions/1768732/do-there-exist-bump-functions-with-uniformly-bounded-derivatives, no such bump function has bounded derivatives, which violates the second condition of the OP. – Jason DeVito - on hiatus Jul 11 '17 at 03:48
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    The claim holds, provided one can show that $f(q) = e^q$ for $q \ge 0$ rational. – Qeeko Jul 11 '17 at 04:40
  • I somehow expect that Gronwall's inequality may be useful here, but I don't know how. – Nick Alger Jul 11 '17 at 08:00
  • @zen This doesn't work. If you differentiate sufficiently many times, $f^{(n)}(x)$ dips below zero for $x>2\pi$; in fact, it dips below zero for arbitrarily large $x$ (for $n$ large enough). Counterintuitively, $g(x) = \frac{\sin(2\pi x)}{2\pi x}$ does not have uniformly bounded derivatives (at least according to Desmos). – MathematicsStudent1122 Jul 11 '17 at 13:48
  • For $x \geq 0$ the solution might be unique, but hardly for $x < 0$ where we don't have much constraints. – md2perpe Jul 11 '17 at 17:37
  • @MathematicsStudent1122 you are right ill delete the comment – zen Jul 11 '17 at 18:01
  • What is assumed about $f$? Are we showing that if $f$ is differentiable on all of $\mathbb{R}$, then $f$ has to be $e^x$? Or do we assume $f$ is continuously differentiable? – mathworker21 Jul 11 '17 at 23:54
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    @mathworker21 $f \in C^{ \infty }$ – Red shoes Jul 11 '17 at 23:58
  • How do you know $f(x)-e^x \in L^1(\mathbb{R})$? – mathworker21 Jul 12 '17 at 01:41

2 Answers2

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We can show that if $$ f^{(n)} \ge 0\qquad{\rm(1)} $$ everywhere, for each integer $n >0$, and $f(x)=e^x$ on at least four points of $\mathbb R$, then $f(x)=e^x$ everywhere. This will make use of Bernstein's theorem. I feel that there must be a more elementary proof though.

I'll start with the simpler case where (1) holds for all $n\ge0$, in which case it is only necessary to suppose that $f(x)=e^x$ at three points. As $\left(-\frac{d}{dx}\right)^nf(-x)=f^{(n)}(-x)\ge0$, by definition $f(-x)$ is completely monotonic. Bernstein's theorem means that we can write $$ f(-x)=\int e^{-xy}\,\mu(dy)\qquad {\rm(2)} $$ for some finite measure $\mu$ on $[0,\infty)$ and for all $x > 0$. By inversion of Laplace transforms, $\mu$ is uniquely defined. In fact, the statement (2) applies for all $x\in\mathbb R$ and we can argue as follows -- Applying the same argument to $f(K-x)$, for any fixed real $K$ extends (2) to all $x-K > 0$ and, letting $K$ go to $-\infty$, to all $x\in\mathbb R$. Changing the sign of $x$, for convenience, $$ f(x)=\int e^{xy}\,\mu(dy). $$ See also, this answer on MathOverflow (and the comments). Suppose that $\mu$ has nonzero weight outside of the set $\{1\}$. Multiplying by $e^{-x}$ and taking second derivatives $$ \left(\frac{d}{dx}\right)^2e^{-x}f(x)=\left(\frac{d}{dx}\right)^2\int e^{x(y-1)}\mu(dy)=\int(y-1)^2e^{x(y-1)}\mu(dy) > 0. $$ This means that $e^{-x}f(x)$ is strictly convex, contradicting the fact that it is equal to 1 at more than two points of $\mathbb R$. So, $\mu$ has zero weight outside of $\{1\}$ and $e^{-x}f(x)=\mu(\{1\})$ is constant.

I'll now return to the case where (1) holds for $n > 0$ and $f(x)=e^x$ on a set $S\subseteq\mathbb R$ of size four. By the mean value theorem, $f^\prime(x)=e^x$ holds for at least one point between any two points of $S$ and, hence, holds for at least three points of $\mathbb R$. So, using the proof above, $f^\prime(x)=e^x$ everywhere. Integrating, $f(x)=e^x+c$ for a constant $c$. Then, in order that $f(x)=e^x$ anywhere, $c$ must be zero.

George Lowther
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  • Your proof applies to $x > 0$, right? – mathworker21 Jul 12 '17 at 04:12
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    It applies to all $x$. Unfortunately, the statement of Bernstein's theorem is for $x > 0$, but the result applies for all $x$ and I tried to give a quick argument why in my proof. – George Lowther Jul 12 '17 at 04:14
  • Very Small nitpick: The Bernstein theorem also requires $n=0$ but the question only gives $\mathbb N^*={1,2,3...}$. – i9Fn Jul 12 '17 at 08:51
  • Good point, I misread that bit. Increasing the number of points to 4, you should still be able to apply a similar argument to $f^\prime$ to get $f^\prime(x)=e^x$, then deduce the result for $f$. – George Lowther Jul 12 '17 at 09:08
  • In Jordan Bell's notes, completely monotone functions are defined to be nonincreasing and have finite limits at $0$ and $\infty$. His proof depends on these properties and only applies for $x \in [0,\infty)$ (his $x$ has the opposite sign from yours). I don't see how making a change of variables so $[0, \infty)$ becomes $[K, \infty)$ and letting $K \to -\infty$ constitutes a proof, particularly in the case when $f$ is unbounded. If your much stronger result could be obtained so easily, I'd expect it would be mentioned. – Keith McClary Jul 14 '17 at 04:45
  • My result is not much stronger. It is a minor extension for the special case where the function is completely monotone on the whole real line and not just the positive reals. The extension is not needed anyway, as you could first prove that $f(x)=e^x$ on $(-\infty,-K]$ and only then taking the limit as $K$ goes to $-\infty$. – George Lowther Jul 14 '17 at 11:20
  • Sorry, I misunderstood. So, your argument (if you start with the sign reversed statement of the Question) is: "We show that this holds for any three distinct points of agreement. It is sufficient to show this for any interval $[K, \infty]$ including the three points. Translating by $-K$ we have the conditions of Bernstein's theorem. ...". – Keith McClary Jul 27 '17 at 03:55
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Perhaps someone can formalize this for me. Because f(n) = e^n some derivative of f(x) must not equal e^x for them to be different functions. If the yth derivative of f(x) > e^x then f(x) > e^x for x > some number unless the y+zth derivative of f(x) < e^x in witch case f(x) < e^x for x > some number. For each derivative where that derivative of f(x) != e^x a higher derivative of f(x) that is > or < e^x must exist to compensate. Each compensation can only make f(x) < or > e^x for x > some number. Thusly f(n) cannot equal e^n for all n unless all derivatives of f(x) equal e^x meaning f(x) = e^x.

user459832
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