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Inspired by the linear independence of the solutions to linear homogeneous constant coefficient ODEs, I want to prove that the set of functions $$S=\left\{z \mapsto z^\alpha \exp(\beta z):\alpha,\beta \in \mathbb{C} \right\} $$ is linearly independent over the complexes. Clearly some care has to be taken with defining $z^\alpha$ for $\alpha \notin \mathbb{Z}$. Here is my attempt at a proof:

Let $f_k(z)=z^{\alpha_k} \exp(\beta_k z)$ for $k=1,\dots,N$ be distinct functions of $S$, for any branch of the powers, and suppose that $$\sum_{k=1}^N c_k f_k(z) \equiv 0. $$ Suppose w.l.o.g. that $\beta_1$ has the largest real part, then division gives $$\sum_{k=1}^N c_k f_k(z) \exp(-\beta _1 z) \equiv 0, $$ and taking the limit as $z \to +\infty$ leaves us only with those powers $z^{\alpha_k}$, whose corresponding $\beta=\beta_1$. That is $$\sum_{k=1}^n c_k z^{\alpha_k} \equiv 0, $$ where $f_k(z)=z^{\alpha_k} \exp(\beta_1 z)$ for $1 \leq k \leq n \leq N$. Using the linear independence of complex powers (which I won't prove here), we find $c_1=c_2=\dots=c_n=0$. The proof can then proceed by moving on to the $\beta$ with the second largest real part.

Is the above correct? Any corrections and/or more elegant proofs are welcome. Thanks!

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  • $\begingroup$ Why didn't you use Wronskian for this family? Should be $W(y_1,y_2)\neq0$. $\endgroup$
    – Nosrati
    Oct 7, 2017 at 14:35
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    $\begingroup$ That proof does not treat the case where more than one $\beta_i$ has the same real part. $\endgroup$
    – Dap
    Oct 7, 2017 at 14:36
  • $\begingroup$ @MyGlasses Doesn't the non-vanishing Wronskian criterion hold only for solution spaces of some linear ODE? I can't find a common ODE for this family. $\endgroup$
    – user1337
    Oct 7, 2017 at 14:39

2 Answers 2

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HINT:

The idea with the real part is OK, but you run into some trouble is two of them have the same real part. Better is to consider a $\beta_k$ of largest absolute value $\beta_k = r_k e^{i \omega_k}$ and then take $z= e^{-i \omega_k} t$, with $t\to \infty$. Assume that we have $$\sum_{l} P_l(z) \exp(\beta_l z) = 0$$ Then we get $$P_k(z) = - \sum_{l \ne k} P_l(z) \exp ( (\beta_l - \beta_k)z)$$

Now, for $z= e^{-i \omega_k} t= t \cdot \mu$, and $t$ large, the RHS decreases like $\exp(- c t)$ for some $c>0$, while the LHS will be $\approx t^\alpha$, for some $\alpha \in \mathbb{R}$, so cannot decrease that fast to $0$.

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Here is a different argument using complex analysis.

First I will treat the case where the $\alpha_k$ are integers. Note we do not need to deal with the branch cut in this case. Suppose there is a linear dependency $$f(z)=\sum_{k} c_{k} z^{n_k} e^{\beta_k z}\equiv 0.$$ By multiplying by $z^{-\min_kn_k}$ we may assume $n_k\geq 0$ for all $k$. The Laplace transform $$F(s)=\int _{0}^{\infty }f(t)e^{-st}\,dt$$ is defined for sufficiently large $\operatorname{Re}(s)$ and a table of standard transforms shows that it is $$F(s)=\sum_{k} c_{k} \frac{\Gamma(n_k+1)}{(s-\beta_k)^{n_k+1}}$$ where $\Gamma$ is the gamma function. This extends to a meromorphic function and the order and coefficients of the poles are determined by the residue theorem. But since $f\equiv 0$ we have $F\equiv 0$, proving that $c_k=0$ for all $k$ as required.

Finally we can reduce to the case where $\alpha_k$ are integers. Consider the linear map $T$ that takes $z^\alpha e^{\beta z}$ to $e^{2\pi i\alpha}z^{\alpha} e^{\beta z}$. This can be defined extensionally as the result of taking the analytic continuation clockwise once around the origin until you end up where you started. By linear independence of characters, this lets us break up any linear dependency into the eigenspaces $S_\alpha$ consisting of the functions $z^{\alpha+n} e^{\beta z}$ with $n\in\mathbb Z$ and $\beta\in\mathbb C$. By multiplying by $z^{-\alpha}$ this reduces to the case where $\alpha$ is an integer.

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