On existence of a suitable density for a sequence

Question

Suppose that $\lambda_1<\lambda_2<\ldots$ is a sequence of positive real numbers such that $$|\{n\in \mathbb N\,:\, \lambda_n \leq \lambda\}| \leq \sqrt{\lambda} \quad \forall\, \lambda>>1.$$ Let $\{a_k\}_{k=1}^\infty\subset \mathbb C$ be a sequence with Schwartz decay for large $k$ (i.e. its elements $a_k$ exhibit decay faster than any polynomial in $k$ for large $k$). Assume that $$ \sum\limits_{k=0}^\infty a_k e^{i \lambda_k x} =0 \qquad \forall\, x\in [0,1].$$ Does it follow that $a_k=0$ for all $k\in \mathbb N$?

Answer 1

The answer is yes. First of all let me consider interval $[-\frac{1}{2},\frac{1}{2}]$ for convenience by multiplying $a_k$ by $e^{\frac{1}{2}i\lambda_k}$. Assume that, say, $a_1\neq 0$. Let us take some $f\in L^2[-\frac{1}{2}, \frac{1}{2}]$ and assume that $$\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) e^{i\lambda_k x}dx = 0, k\ge 2.$$

We obviously have $$\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \sum_{k=1}^\infty a_k e^{i\lambda_k x}dx = 0,$$

because the sum is zero by assumption. Since $a_k\in \ell^1$, the sum converges absolutely and uniformly and hence in $L^2[-\frac{1}{2}, \frac{1}{2}]$, so we can switch summation and integration. After this switch only $k = 1$ survives and we get $$a_1\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x)e^{i\lambda_1 x}dx = 0,$$

so $\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x)e^{i\lambda_1 x}dx = 0$. Technically, here I can just construct a function $f$ for which this is false, but let me before that do one more thing which will hopefully clarify the matter and explain our construction of $f$. Let us take the Fourier transform. Then from $\hat{f}(-\lambda_k) = 0, k\ge 2$ we conclude that $\hat{f}(-\lambda_1) = 0$. But $\hat{f}$ is an entire function by the (easy part of the) Paley--Wiener theorem.

Consider $\hat{g}(x)=\frac{\hat{f}(x)}{x+\lambda_1}$. It is still an entire function and by the (hard part of the) Paley--Wiener theorem $g\in L^2[-\frac{1}{2}, \frac{1}{2}]$ (although here there is a way to see that $g\in L^2[-\frac{1}{2},\frac{1}{2}]$ without ever doing the Fourier transform). But $g(-\lambda_k), k\ge 2$ is still zero! So by applying our previous reasoning $g(-\lambda_1) = 0$. Repeating this indefinitely $\hat{f}$ has zero of infinite order at $-\lambda_1$, but since it is an entire function, $\hat{f}$ is just identically zero, so $f$ is identically zero. Therefore, $e^{i\lambda_k x}$ is a complete system in $L^2[-\frac{1}{2}, \frac{1}{2}]$, so we just have to construct a non-zero function for which $\int_{\frac{1}{2}}^{\frac{1}{2}} f(x)e^{i\lambda_k x}dx = 0$.

Now, first let us construct some entire function vanishing on $-\lambda_k$. Well, this is what Weierstrass' product is for! Consider $$h(z) = \prod_{k=1}^\infty \left(1 + \frac{z}{\lambda_k}\right)$$

(and if one of $\lambda_k$ is zero then replace the corresponding factor by just $z$). By elementary estimates (e.g. taking logarithm and doing summation by parts) we can show that the product converges to a non-zero entire function $h$ with $|h(z)|\le C e^{C\sqrt{|z|}}$ for some absolute constant $C$. The key fact about this growth rate is that the logarithmic integral for it converges, meaning $$\int_{-\infty}^\infty \frac{\log(Ce^{C\sqrt{|x|}})}{1+x^2}dx < \infty.$$

Our next ingredient is the Beurling--Malliavin theorem. Again, in general it is a highly non-trivial theorem, although for monotone weights (like in our case) it has a simpler proof. From it we can find a function $g\in L^2[-\frac{1}{2},\frac{1}{2}]$ with $|\hat{g}(x)|\le Ce^{-2C\sqrt{|x|}}, x\in \mathbb{R}$, say. Consider $f$ with $\hat{f}(x)=\hat{g}(x)h(x)$. Note that because we killed the growth of $h$ by the rapid decay of $\hat{g}$, in particular it is easy to see that $\hat{f}$, and hence $f$, is in $L^2(\mathbb{R})$. Since we have $h(x)$ factor we have $\hat{f}(-\lambda_k) = 0$. It remains to prove that $f\in L^2[-\frac{1}{2}, \frac{1}{2}]$ actually and not just in $L^2(\mathbb{R})$.

For this we once again invoke the (hard part of the) Paley--Wiener theorem, and this time while there is a way to circumvent the reference to the general theorem, it is a bit harder. Let me (finally!) recall its statement: if the function $F(z)$ is an entire function which is in $L^2(\mathbb{R})$ and which satisfies the bound $|F(z)|\le C_\varepsilon e^{(\frac{1}{2}+\varepsilon)|z|}$ for all $\varepsilon > 0$ then $\hat{F}$ is supported on $[-\frac{1}{2}, \frac{1}{2}]$.

Our function is $\hat{g}(z)h(z)$. We already showed that it is in $L^2(\mathbb{R})$. For $\hat{g}(z)$ we have a bound (obtainable directly from the Fourier transform formula) $|\hat{g}(z)| \le Ce^{\frac{1}{2}|z|}$, while for $h(z)$ we have a bound $|h(z)|\le Ce^{C\sqrt{|z|}}$, which can be easily seen to be at most $C_\varepsilon e^{\varepsilon |z|}$ for all $\varepsilon > 0$. So for the function $\hat{g}(z)h(z)$ we proved the required bound, hence its Fourier transform (which if $f$) is in $L^2[-\frac{1}{2}, \frac{1}{2}]$, as required.

Since it might not be obvious behind all the technical details, let me emphasize that the cornerstone fact is that for the function $h$ the logarithmic integral of its majorant converges. This is exactly where we used the assumption about the growth of $\lambda$'s. Note that just assuming that $\lambda$'s have zero density would not be enough, we need something stronger (for example $\lambda^{1-\varepsilon}$ would still work, but say $\frac{\lambda}{\log \lambda}$ probably wouldn't).