Boundedness of functional In the setting of $2\pi$-periodic $C^1$ functions (whose Fourier series converge to themselves), and given a linear functional $D:C^1_{\text{per}}\to\mathbb R$ satisfying $\sup_{n}|D(e^{inx})|<\infty$ I would like to show that $D$ is continuous (or equivalently, bounded).
Attempt
The supremum condition seems it should imply boundedness, but I'm not managing to formalise that and keep running into a circular argument. For example, by contradiction, let's assume a series $\{f_n\}$ where $\|f_n\|=1$ such that $D(f_n) > n^2$, then taking $\frac {f_n}n$ which has norm $\frac 1n$ we have $D(\frac {f_n}n) > n$ which is a contradiction if $D$ is continuous, but that's what we're trying to prove...
Cheers.
Edit
I think in $L^2$ the hypothesis is wrong since we can construct a sequence 
\begin{align}
f_1 &= e_1 \\
f_2 &= \frac 1 {\sqrt 2}e_1 + \frac 1{\sqrt 2}e_2 \\
&\ \vdots \\
f_n &= \sum_{k=1}^n \frac 1{\sqrt n}e_k
\end{align}
So $\|f_n\|_2^2 = 1$ but if $D(e_k)=M$ (which satisfies our bound condition) we have $D(f_n) = \frac {nM}{\sqrt n}\to \infty$
In light of the failure of the $L^2$ norm, I suppose the $C^1$ norm $\|f\|_\infty+\|f'\|_\infty$ is the focus here.

This is not true.  The functions $\{e^{inx}\}$ do not span $C^1_{\text{per}}$; indeed, no countable set can.  So using the axiom of choice, one can show the existence of a linear functional that vanishes on all the $e^{inx}$ but is not continuous.
You can't tell whether a linear functional is continuous by looking at a proper subspace, even if that subspace is dense.
You can, however, show that there exists a unique continuous linear functional $D_1$ such that $D_1(e^{inx}) = D(e^{inx})$ for every $n$.  Is that what you want?
