Lacunary Fourier series and Hölder continuity at a point on the circle Let $(\lambda_{n})$ be lacunary (i.e. $\exists$ constant $q>1$ such that $\lambda_{n+1}>q\lambda_{n}$ for all $n\in\mathbb{N}$); $f\in L^{1}(T)$ with Fourier series $\sum_{n\in\mathbb{N}}a_{n}\cos(\lambda_{n}t)$ and suppose also that $f$ is Hölder continuous at a point $t_{0}\in T$.
I want to show that $a_{n}=O(\lambda^{-\alpha}_{n})$ and also that $f$ is Hölder continuous on $T$.
Thus we have that $\hat{f}(n)=\sum_{n\in\mathbb{N}}a_{n}\cos(\lambda_{n}t)$ and that $\exists C>0$ such that $|f(x+t_{0})-f(t_0)|\le C|x|^{\alpha}$, but I do not know how to proceed with tackling the problem.
 A: First, your formula $\hat{f}(n)=\sum_{n\in\mathbb{N}}a_{n}\cos(\lambda_{n}t)$ is way off. In fact $\hat f(\pm\lambda_n)=a_n/2$ and $\hat f(j)=0$ if $j\ne\lambda_n$.
Note as well that assuming $f$ is $Lip_\alpha$ at one point implies $a_n=O(\lambda_n^{-\alpha})$ for $0<\alpha\le 1$, but that condition on the $a_n$ only implies that $f\in Lip_\alpha$ for $0<\alpha<1$. (For $\alpha=1$ it implies that $f$ is in the so-called Zygmund class.)
First Half:
We need to construct a few kernels. The construction is actually easier on the line, where we can use dilations; we'll use Poisson summation to get from the line back to the circle. (The reason I'm writing this out instead of just referring to some book is that people often construct the required kernels as linear combinations of Fejer kernels; I feel the construction below is much more transparent.)
Choose a Schwarz function $\psi$ with $$\hat\psi(1)=1$$  and $$supp(\hat\psi)\subset\left(\frac1q,q\right).$$  For $\lambda>0$ let $$\psi_\lambda(t)=\lambda\psi(\lambda t),$$so that $||\psi_\lambda||_1=||\psi||_1$ and $$\hat\psi_\lambda(\xi)=\hat\psi\left(\frac\xi\lambda\right);$$in particular $\hat\psi_\lambda(\lambda)=1$ and $\hat\psi_\lambda$ is supported in $(\lambda/q,\lambda q)$.
Now define a $2\pi$-periodic function $\phi_n$ by $$\phi_n(t)=\sum_{j\in\Bbb Z}\psi_{\lambda_n}(t+2\pi j),$$so that $$||\phi_n||_{L^1(\Bbb T)}\le\||\psi_{\lambda_n}||_{L^1(\Bbb R)}=||\psi||_{L^1(\Bbb R)}$$and $$\hat\phi_n(k)=\widehat{\psi_{\lambda_n}}(k)\quad(k\in\Bbb Z).$$(There are a few $\pi$'s missing here; look up Poisson summmation and get this straight before you hand it in.) See Details below.
In particular $$\hat\phi_n(\lambda_n)=1$$and $$\hat\phi_n(k)=0\quad(k\notin(\lambda_n/q,\lambda_nq),$$so $\hat\phi_n(\lambda_j)=0$ if $j\ne n$.
Assume $t_0=0$ for convenience. Since $f$ has period $2\pi$ the definition of $\phi_n$ shows that we have $$a_n=\frac1\pi\int_{-\pi}^\pi f(t)\phi_n(-t)\,dt=\frac1\pi\int_{-\infty}^\infty f(t)\psi_{\lambda_n}(-t)\,dt
=\frac1\pi\int_{-\infty}^\infty f(t/\lambda_n)\psi(-t)\,dt.$$ 
Now since $\int\psi=0$ it follows that 
$$a_n=
\frac1\pi\int_{-\infty}^\infty (f(t/\lambda_n)-f(0))\psi(-t)\,dt.$$
So $$|a_n|\le\lambda_n^{-\alpha}\frac C\pi\int_{-\infty}^\infty|t|^\alpha|\psi(t)|\,dt=O(\lambda_n^{-\alpha}).$$
Second Half:
Now to show that that implies $f\in Lip_\alpha$, if $0<\alpha<1$: We have $$|f(s)-f(t)|\le c\sum\lambda_n^{-\alpha}|e^{i\lambda_ns}-e^{i\lambda_nt}|.$$Note that $$|e^{i\lambda s}-e^{i\lambda t}|\le 2$$and also $$|e^{i\lambda s}-e^{i\lambda t}|\le \lambda|s-t|.$$We use one estimate on some of the terms and the other on the other terms. Choose $N$ so $\lambda_N|s-t|\le 1$ and $\lambda_{N+1}|s-t|>1$. Then $$\sum_{n=1}^N\lambda_n^{-\alpha}|e^{i\lambda_ns}-e^{i\lambda_nt}|
\le\sum_{n=1}^N\lambda_n^{1-\alpha}|s-t|
\le\lambda_N^{1-\alpha}|s-t|\sum_{k=-\infty}^0q^{k(1-\alpha)}
=c\lambda_N^{1-\alpha}|s-t|^{1-\alpha}|s-t|^\alpha\le c|s-t|^\alpha$$
(note we used $1-\alpha>0$ there, twice.)
On the other hand, since $\alpha>0$ we have $$\sum_{n=N+1}^\infty\lambda_n^{-\alpha}|e^{i\lambda_ns}-e^{i\lambda_nt}|
\le 2\sum_{n=N+1}^\infty\lambda_n^{-\alpha}
\le 2\lambda_{N+1}^{-\alpha}\sum_{k=0}^\infty q^{-k\alpha}
=c\lambda_{N+1}^{-\alpha}\le c|s-t|^\alpha.$$
Details The essence of the Possion Summation Formula is this:
Proposition Suppose $f\in L^1(\Bbb R)$. Then the series $$g(t)
=\sum_{j\in\Bbb Z}f(t+2\pi j)$$converges absolutely almost everywhere. We have $$||g||_{L^1(\Bbb T)}\le\frac1{2\pi}||f||_{L^1(\Bbb R};$$further, if $n\in\Bbb Z$ then $$\hat g(n)=\frac1{2\sqrt\pi}\hat f(n).$$
Proof: Monotone convergence, the translation invariance of Lebesgue measure and then MCT again show that $$\frac1{2\pi}\int_0^{2\pi}\sum_{j\in\Bbb Z}|f(t+2\pi j)|
=\frac1{2\pi}\sum_{j\in\Bbb Z}\int_{2\pi j}^{2\pi(j+1)}|f(t)|\,dt
=\frac1{2\pi}\int_{-\infty}^\infty|f(t)|\,dt.$$So the sum converges almost everywhere. If $n\in\Bbb Z$ then $e^{-int}$ has period $2\pi$; since the sum converges in $L^1$ this shows that $$\hat g(n)=\frac1{2\pi}\int_0^{2\pi}g(t)e^{-int}
=\frac1{2\sqrt\pi}\frac1{\sqrt\pi}\int_{-\infty}^\infty f(t)e^{-int}
=\frac1{2\sqrt\pi}\hat f(n).$$
