If a periodic function with lacunary Fourier series is zero on a small interval then is it smooth? Let $f\in L^{1}_{2\pi}=L(T)$ with $f\sim\sum a_{n}\cos(\lambda_{n}t)$ with $(\lambda_{n})$ lacunary and $f$ Hölder continuous on the circle $T$. Moreover, $a_{n}=O(\lambda^{-\alpha}_{n})$.
Suppose that $f=0$ on a small interval in $T$. I want to show that $f\in C^{\infty}(T)$
I would appreciate some hints on how to solve this question. I know that the smoother a function the faster the Fourier series will converge and from the Riemann-Lebesgue lemma we know that $a_{n}\to 0$ as $n\to\pm\infty$.
Thus for arbitrary $\epsilon>0$ our assumption is that 
$$f=
0\mbox{ on } [-\epsilon,\epsilon]\subset T=[0,2\pi]$$
i.e. $\text{supp }f\subset T\setminus[-\epsilon,\epsilon]$.
But I am struggling to build the implication. I have a feeling that this is quite basic though and I have put more information than is necessary.
 A: It indeed follows that $f\equiv 0$, as already suspected by David in the comments.
To see this, consider $g=\sum a_n z^{\lambda_n}$. This is a holomorphic function on the unit disk. Notice that $f(t)=2\textrm{Re}\, g(e^{it})$, so your assumption says that $g$ is purely imaginary on an arc, and this lets us holomorphically continue $g$ across this arc, by reflection.
However, it is well known that lacunary power series can not be extended past the unit circle. See here.
A: My first reaction was that assuming $f$ vanishes on an interval must imply that $f=0$, and assuming just that $f$ is smooth on an interval should imply that $f$ is smooth.
In fact (i) if $f$ is infinitely differentiable at just one point of the boundary then $f$ is smooth and (ii) yes, if $f$ vanishes on an interval then $f=0$. These are both special cases of the general principle that the behavior of a lacunary series is the same everywhere. We give a proof of (i) below. Another answer has given a one-line proof of (ii), based on a certain well-known fact about lacunary series; below we indicate how one can prove (ii) starting from nothing.
A natural question given all this is whether assuming just that $f$ is infinitely differentiable at one point and every derivative vanishes at that point implies $f=0$. I don't know, although I suspect not, since smoothness is far from real-analyticity.
(The hypothesis that $f$ is Holder continuous is redundant, by the way; if $f$ vanishes on an interval then $f$ certainly satisfies a Holder condition at one point, so the result here shows that $f\in Lip_\alpha$.)
Proof of (i)
Assume that $f$ is infinitely differentiable at the origin, and note that that previous result implies that $f$ is continuous, hence bounded.
Recall that there exists a Schwarz function $\psi$ such that $\hat\psi$ vanishes in a neighborhood of the origin and $$a_n=
\frac1\pi\int_{-\infty}^\infty\psi(-t)f(t/\lambda_n)\,dt.$$Now $D^k\hat\psi(0)=0$ for all $k$, hence $$\int\psi(-t)t^k\,dt=0$$for all $k$. So if $P_k$ is the Taylor polynomial for $f$ of order $k$ we have $$a_n=
\frac1\pi\int_{-\infty}^\infty\psi(-t)(f(t/\lambda_n)-P_k(t/\lambda_n))\,dt.$$
Now there is a version of Taylor's theorem valid for functions $k$-times differentiable at just one point, showing that $$f(t)-P_k(t)=o(|t|^k)\quad(t\to0).$$ Since $f$ is bounded this shows that $$|f(t)-P_k(t)|\le c|t|^k\quad(t\in\Bbb R).$$So we have $$|a_n|\le C_k\lambda_n^{-k}\int|\psi(-t)|\,|t|^k=O(\lambda_n^{-k}).$$So $f$ is smooth.
Partial Proof of (ii)
Another useful rule of thumb when dealing with lacunary series: The case $\lambda_n=2^n$ is probably easier to prove, and if it's true in that case it's true in general, sometimes by a generalization of the proof for the special case. We prove (ii) for $\lambda_n=2^n$ and then indicate how one might extend the proof.
So assume that $\lambda_n=2^n$ and that $f$ vanishes on an interval $I$. Note that it follows that $f$ is smooth, by (i).
Choose $N$ so $2\pi2^{-N}<|I|$; now replace $I$ by a subinterval so that $$|I|=2\pi2^{-N}.$$ Letting $f^{(k)}$ denote the $k$-th derivative, write $$f^{(k)}=\sum_{n=1}^\infty=\sum_{n=1}^N+\sum_{n=N+1}^\infty=f_1^{(k)}+f_2^{(k)}.$$Since $f_1^{(k)}+f_2^{(k)}$ vanishes on $I$ we have $$\frac{1}{|I|}\int_I|f_1^{(k)}|^2=\frac1{|I|}\int_I|f_2^{(k)}|^2.$$The triangle inequality shows that $$\frac{1}{|I|}\int_I|f_1^{(k)}|^2\le c_1\lambda_N^{2k}$$(here $c_1$ depends on $N$ and on the coefficients, but is independent of $k$; similarly for $c_2$ below)
And now here's the part where $\lambda_n= 2^n$ is convenient: The interval $I$ is actually a period of $f_2^{(k)}$, so $$\frac{1}{|I|}\int_I|f_2^{(k)}|^2=\lVert f_2^{(k)}\rVert_2^2.$$
Assume $f_2\ne0$. Then there exists $c_2>0$ so that $$\frac{1}{|I|}\int_I|f_2^{(k)}|^2\ge c_2\lambda_{N+1}^{2k}.$$So we have $$c_2\lambda_{N+1}^{2k}\le c_1\lambda_{N}^{2k}$$for all $k$, which is impossible (since $c_2>0$).
So $f_2=0$. Hence $f_1=0$ on $I$, which implies that $f_1=0$ by any number of arguments (for example $f_1(t)$ is a rational function of $z=e^{it}$). QED.
Comment on (ii) in General
The same proof works for a general lacunary series if we can show that $$\frac{1}{|I|}\int_I|f_2^{(k)}|^2\ge c\lVert f_2^{(k)}\rVert_2^2.$$And we can, although the details get a little technical.
In the special case $\lambda_n=2^n$ this was automatic because the functions $(\cos(\lambda_nt))_{n>N}$ were actually orthonormal on $I$. Now they're no longer orthonormal, but they're close to orthonormal, meaing that the inner products on $I$ are small. One can use the almost-orthonormality to show that the inequality holds.
