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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.

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  • $\begingroup$ 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, 2017 at 12:27
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    $\begingroup$ What is $\Bbb N^*$? $\Bbb N$ with zero? $\endgroup$
    – M. Winter
    Jun 13, 2017 at 14:59
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    $\begingroup$ 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, 2017 at 3:34
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    $\begingroup$ The claim holds, provided one can show that $f(q) = e^q$ for $q \ge 0$ rational. $\endgroup$
    – Qeeko
    Jul 11, 2017 at 4:40
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    $\begingroup$ @mathworker21 $f \in C^{ \infty }$ $\endgroup$
    – Red shoes
    Jul 11, 2017 at 23:58

2 Answers 2

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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.

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  • $\begingroup$ Your proof applies to $x > 0$, right? $\endgroup$ Jul 12, 2017 at 4:12
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    $\begingroup$ 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, 2017 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, 2017 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, 2017 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, 2017 at 4:45
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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.

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