Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Leont'ev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

  • $\begingroup$ I would be surprised if the answer is yes (but also very pleased to see such a wonderful theorem!), since local uniform convergence on $\mathbb{C}$ is much, much stronger than on $\mathbb{R}$. However, the special form of exponentials compensates a lot for lack of complex analytic contour integration trickery; so anyway, I'm convinced that the question is interesting! I'm aiming for a counterexample though...! $\endgroup$
    – Zen Harper
    Jul 16, 2010 at 16:41
  • $\begingroup$ Zen, thank you for the comment. I have a similar gut feeling about this. $\endgroup$ Jul 16, 2010 at 16:57
  • 1
    $\begingroup$ In the original message, "A. F. Lavrent'ev" should be "A. F. Leont'ev". $\endgroup$ Aug 4, 2012 at 11:06
  • $\begingroup$ @Alexandre Eremenko: Many thanks! I stand corrected. $\endgroup$ Aug 5, 2012 at 9:03

3 Answers 3


I have some partial answers.

I. It is not hard to construct a Dirichlet series $$f(z)=\sum_{n=1}^\infty a_ne^{\lambda_n z}$$ which converges to $0$ absolutely and uniformly on the real line but does not converge at some points of the complex plane. It is constructed as a sum of 3 series $f=f_0+f_1+f_2.$ Let $f_1$ be a series with imaginary exponents $\lambda_n$ which converges to an entire function in the closed lower half-plane, but not in the whole plane. Such series is not difficult to construct, see V. Bernstein, page 34, (see the full reference below) and there are simpler examples, with ordinary Dirichlet series. Then put $f_2=\overline{f_1(\overline{z})}$, and $f_0=-f_1-f_2$. So all three functions are entire. Now, according to Leontiev, EVERY entire function can be represented by a Dirichlet series which converges in the whole plane. Thus we have a Dirichlet series $f_0+f_1+f_2$ which converges on the real line to $0$ but does not converge in the plane.

A counterexample to the original question also requires real coefficients, this I do not know how to do (for $f_0$).

II. It is clear from the work of Leontiev, that to obtain a reasonable theory, one has to restrict to exponents of finite upper density, $n=O(|\lambda_n|)$, otherwise there is no uniqueness in $C$. In the result I cited above the expansion of $f_0$ is highly non-unique.

Assuming finite upper density I proved that if a series is ABSOLUTELY and uniformly convergent on the real line to zero, then all coefficients must be zero. http://www.math.purdue.edu/~eremenko/dvi/exp2.pdf I don't know how to get rid of the assumption of absolute convergence.

But there is a philosophical argument in favor of absolute convergence: the notion of "linear dependence" should not depend on the ordering of vectors:-)

III. The most satisfactory result on my opinion, is that of Schwartz. Let us say that the exponentials are S-linearly independent if none of them belongs to the closure of linear span of the rest. Topology of uniform convergence on compact subsets of the real line. Schwartz gave a necessary and sufficient conditon of this: the points $i\lambda_k$ must be contained in the zero-set of the Fourier transform of a measure with a bounded support in R.

(L. Schwartz, Theorie generale des fonctions moyenne-periodiques, Ann. Math. 48 (1947) 867-929.)

A complete explicit characterization of such sets is not known, but they have finite upper density, and many of their properties are understood. These Fourier transforms are entire functions of exponenitial type bounded on the real line. The link I gave above contains Schwartz's proof in English. S-linear dependence is also non-sensitive to the ordering of functions, which is good.

IV. Vladimir Bernstein's book is "Lecons sur les progress recent de la theorie des series de Dirichlet", Paris 1933. This is the most comprehensive book on Dirichlet series, but unfortunately only with real exponents.

V. The application to the functional equation mentioned by the author of the problem is not a good justification for the study of the problem in such generality. The set of exponentials there is very simple, and certainly we have $R$-linear independence for SUCH set of exponentials. Besides the theorem stated as an application has been proved in an elementary way.

VI. Finally, I recommend to change the definition of $R$-linear independence by allowing complex coefficients (but equality to $0$ on the real line). Again in the application mentioned in the original problem, THIS notion of $R$-uniqueness is needed: the function is real, but the exponentials are not real, thus coefficients should not be real.

  • $\begingroup$ In the original text of my answer, the numeration of sections is natural: 1, 2, 3, 4, 5,... Why the web site produces something weird, I do not know. $\endgroup$ Oct 1, 2012 at 21:09

EDIT: The following actually assumes uniform absolute convergence on compact regions, and thus addresses a slightly different problem, as Zen Harper points out in the comments.

This is an observation which is too big to fit in comments. I claim that if there exists $M>0$ such that for all $n$, $|\Im \lambda_n| < M$, then the answer is yes ($\mathbb{R}$-independence implies $\mathbb{C}$-independence). In this case it's easy to see that uniform convergence on compact intervals of $\mathbb{R}$ implies uniform convergence on compact regions of $\mathbb{C}$.

To see this, write $\lambda_n=x_n+y_ni$ where $x_n, y_n$ are real; writes $s=c+di$. Then $$\sum |a_ie^{\lambda_i s}|=\sum |a_i|e^{x_nc-y_nd}$$ which can be compared to $$\sum |a_i| e^{x_n c}.$$

But then uniform convergence on compact regions of $\mathbb{C}$ implies that the limit function $f$ is holomorphic, and it vanishes identically on $\mathbb{R}$, so $f$ must be identically zero. But then applying $\mathbb{C}$-independence, we have that all the $a_i=0$.

This points in the direction of a counterexample when the $y_i$ are unbounded---if we let the $|y_i|$ tend to $\infty$ rapidly this argument fails dramatically.

  • $\begingroup$ Am I being stupid? Aren't you assuming the stronger property of uniform ABSOLUTE convergence (which is still interesting, but a different question), rather than just uniform convergence, on compact sets? Or is it obvious that they're equivalent in this example? $\endgroup$
    – Zen Harper
    Jul 22, 2010 at 23:23
  • $\begingroup$ Ah indeed I am. Oops. $\endgroup$ Jul 22, 2010 at 23:50

Still thinking about the interesting question!

Not an answer, but too big for a comment.

To show what I meant in my comment to Daniel Litt's answer about the difference between uniform absolute convergence and ordinary uniform convergence:

I think (but it was a while ago when I thought about it) that there exist $u_0, u_1, \ldots$ such that $\sum_{n=0}^\infty u_n z^n$ converges uniformly on the set $\{ z \in \mathbb{C} : |z| \leq 1 \}$, but not uniformly absolutely.

Thus $\sum_{n=0}^\infty |u_n| = +\infty$, but also $$ \forall \, \epsilon>0, \quad \exists \, N(\epsilon)>0 \quad \text{such that}: \quad \forall \, |z| \leq 1, \quad \forall \, m \geq n > N(\epsilon), \quad \left| \sum_{k=n}^{m} u_k z^k \right| < \epsilon. $$ This function $f(z) = \sum_{n=0}^\infty u_n z^n$ satisfies $$ f \in A(D) \setminus W_+(D), $$ which shows in particular that the disc algebra $A(D)$, consisting of functions continuous on the closed unit disc and analytic on the open unit disc, is strictly larger than the Wiener algebra $W_+(D)$ of power series absolutely convergent on the closed unit disc.

So the problem is going to be pretty hard because we can't use absolute convergence (unless I'm being stupid or there is a clever trick exploiting the special structure of the exponential functions).

The easiest "solution" is just to ignore it (i.e. assume absolute convergence!) This gives a slightly different, but still interesting, question.


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