Let $f(x)$ continuous function on $[-\pi,\pi]$ and $ c_n:=\int _{-\pi}^{\pi} f(x)\frac{e^{-inx}}{\sqrt{2\pi}}dx$. Suppose $f(-\pi)=f(\pi)$ and for all $x_0 \in [-\pi, \pi]$, there is $r>0$ s.t., $\forall x \in [-\pi, \pi]$ with $|x-x_0| \leq r, f(x)=\sum _{m=0}^{\infty} a_m(x-x_0)^m$
Then, there exists $K >1$ and constant $C$ s.t., $|c_n|\leq \frac{C}{K^{|n|}}$
I guess I use Cauchy's integral expression, but I cannot take $K>1$.
Given the $2\pi$-periodicy of $f$ and the condition $f(0)=f(2\pi)$, it is well defined a continuous function $$F :S^1\rightarrow\mathbb{C}, e^{it}\mapsto f(t),$$ where $S^1$ is the unit circle in the complex plane. Then the condition $$\forall x_0\in[-\pi,\pi], \exists r>0, \exists(a_n)_{n\in\mathbb{N}}\subset\mathbb{C}, \forall x\in[-\pi,\pi],\left({\left(|x-x_0|\le r\right)\implies \left(f(x)=\sum_{m=0}^\infty a_m(x-x_0)^m\right)}\right) $$ can be worked out to $$\forall z_0\in S^1, \exists r>0, \exists(a_n)_{n\in\mathbb{N}}\subset\mathbb{C}, \forall z\in S^1\cap D_r(z_0), F(z)=\sum_{m=0}^\infty a_m(z-z_0)^m.$$
So, for each point of the unit circle, there exists a ball centered in that point where $F$ coincides with an holomorphic function. By the identity principle for holomorphic functions (or, say, analytic continuation), you can extend $F$ to an holomorphic function defined on an open set surrounding the unit circle. So there exists an annulus of convergence $A_{r,R}$ with $r<1<R$ where the Laurent series of $F$ centered in $0$ converges, say: $$\forall z\in A_{r,R}, F(z)=\sum_{n=-\infty}^\infty c_nz^n.$$ Then the series $$\sum_{n=0}^\infty c_nz^n$$ has a radius of convergence of at least $R$, while the series $$\sum_{n=0}^\infty c_{-n}z^n$$ has a radius of convergence of at least $\frac{1}{r}$. Define $H:=\min(R,\frac{1}{r})$. Then: $$1\ge\limsup_{|n|\rightarrow\infty}(|c_n|H^{|n|})^{1/|n|}=H\limsup_{|n|\rightarrow\infty}|c_n|^{1/|n|}$$ So: $$\limsup_{|n|\rightarrow\infty}|c_n|^{1/|n|}\le \frac{1}{H}.$$ Then, if $1<K<H$, easily follows that there exists a constant $C>0$ such that $$\forall n\in\mathbb{Z}, |c_n|\le\frac{C}{K^{|n|}}.$$ Finally, noticing that the coefficients $(c_n)_{n\in\mathbb{Z}}$ are also the Fourier coefficients of $f$, you have got the desidered result.