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.

  • $\begingroup$ What do you mean by $f^{(y)}(x)>0~\forall y\in\mathbb R,y\ge1$? Particularly, when $y\notin\mathbb N$. $\endgroup$ Jun 10 '17 at 12:27
  • 5
    $\begingroup$ What is $\Bbb N^*$? $\Bbb N$ with zero? $\endgroup$
    – M. Winter
    Jun 13 '17 at 14:59
  • 2
    $\begingroup$ 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. $\endgroup$ Jul 11 '17 at 3:34
  • 1
    $\begingroup$ The claim holds, provided one can show that $f(q) = e^q$ for $q \ge 0$ rational. $\endgroup$
    – Qeeko
    Jul 11 '17 at 4:40
  • 2
    $\begingroup$ @mathworker21 $f \in C^{ \infty }$ $\endgroup$
    – Red shoes
    Jul 11 '17 at 23:58

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.

  • $\begingroup$ Your proof applies to $x > 0$, right? $\endgroup$ Jul 12 '17 at 4:12
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 12 '17 at 4:14
  • $\begingroup$ Very Small nitpick: The Bernstein theorem also requires $n=0$ but the question only gives $\mathbb N^*=\{1,2,3...\}$. $\endgroup$
    – i9Fn
    Jul 12 '17 at 8:51
  • $\begingroup$ 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 $f$. $\endgroup$ Jul 12 '17 at 9:08
  • $\begingroup$ 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. $\endgroup$ Jul 14 '17 at 4:45

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.