Suppose that $f\in C_{2}(\mathbb{R}/2\pi\mathbb{Z})$ and $x_{n} = \sum\limits_{j=0}^{n}e_{j}$, where $e_{j} = \dfrac{\cos jx}{\sqrt{\pi}}$ for $j>0$ and $\dfrac{1}{\sqrt{2\pi}}$ for $j=0$. Prove that $\langle f,x_{n}\rangle\to \sqrt{2\pi}f(0)$.
Since $f''$ is continuous then on unit circle (which is a compact) it is bounded, so $f'' = \sum\limits_{n\in\mathbb{Z}}a_{n}(n^2e_{n})\Rightarrow |a_{n}|\leq \dfrac{C}{n^2}$, this implies that Fourier series for $f(x)$ converge absolutely and uniformly. But then this series converge pointwise (at $0$, for example), so $f(0) = \sum\limits_{j\in\mathbb{Z}}a_{j}e_{j}$, but note that $e_{j}=0,\;j<0$, $e_{0} = \dfrac{1}{\sqrt{2\pi}}$ and $e_{j}=\dfrac{1}{\sqrt{\pi}},\;j>0$, so $f(0) = \dfrac{a_{0}}{\sqrt{2\pi}}+\dfrac{1}{\sqrt{\pi}}\sum\limits_{j=1}^{\infty}a_{j}$. However, $\sum\limits_{j=0}^{n}a_{j} = \sum\limits_{j=0}^{n}\langle f,e_{j}\rangle = \langle f,x_{n}\rangle$, so $\dfrac{1}{\sqrt{\pi}}\langle f,x_{n}\rangle\to f(0)-\dfrac{a_{0}}{\sqrt{2\pi}}$... I must be tired, so can be missing something. Please, help :)