Plancherel's Theorem
Let $f$ $\in$ $(C^{2\pi}$, $||f||_{2}$) where $||f||_{2} = \Big(\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^{2}dx\Big)^{\frac{1}{2}}$. \begin{align*} ||f-S_{n}(f)||_{2} \overset{n\to\infty}{\longrightarrow} 0 \end{align*} where $S_n(f)$ is the Fourier Partial Sum of $f$.
Here is my proof so far however, I know that it must be incorrect because it would imply that $S_{n}(f)$ would converge uniformly to $f$ in the Sup norm. Note that $T_{n}$ is the set of Trig polynomials of degree at most $n$.
\begin{align*} || f - S_{n}(f)||_{2}^{2} &= \frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x) - S_{n} (f)(x)|^{2}dx \\ &\leq \frac{1}{2\pi}\int_{-\pi}^{\pi}||f - S_{n}(f)||_{\infty}^{2}dx \\ &= ||f - S_{n}(f)||_{\infty}^{2}\frac{1}{2\pi}\int_{-\pi}^{\pi}dx \\ &= ||f - S_{n}(f)||_{\infty}^{2}\frac{1}{2\pi}2\pi \\ &= ||f - S_{n}(f)||_{\infty}^{2} \\ &\leq ||f - T_{n}||_{\infty}^{2} \\ \end{align*}
Hence, \begin{align*} || f - S_{n}(f)||_{2}^{2} \leq ||f - T_{n}||_{\infty}^{2} \\ \end{align*}
By Weierstrass's 2nd Approximation Theorem,
\begin{align*} ||f - T_{n}||_{\infty}^{2} \overset{n\to\infty}{\longrightarrow} 0 \implies || f - S_{n}(f)||_{2}^{2} \overset{n\to\infty}{\longrightarrow} 0 \end{align*}
However, this would also imply (with respect to the Sup norm)
\begin{align*} ||f - T_{n}||_{\infty}^{2} \overset{n\to\infty}{\longrightarrow} 0 \implies ||f - S_{n}(f)||_{\infty}^{2} \overset{n\to\infty}{\longrightarrow} 0 \end{align*}
And I know this to be incorrect for any $f\in C^{2\pi}$.