Why space of $2\pi$ periodic continuous functions over $R$ with convolution operation does not have identity? Let $E$ be the space of continuous $2\pi$ periodic functions from real number to real number. Define commutative product on $E$ by $f\star g=1/(2\pi)\int_{-\pi}^\pi f(t)g(x-t)dt$ and it is clear that $f\star g\in E$ whenever $f,g\in E$. Then $E$ becomes commutative Banach algebra with sup norm.
The book asserts that $E$ as Banach algebra does not have identity.(i.e. $\not\exists e\in E$ s.t. $e\star f=f\star e=f$.) 
$\textbf{Q':}$ The book hints to use Stone-Weierstrass theorem. First $E$ is clearly closed by uniform norm. In my line of thought, those functions are essentially functions defined on $[-\pi,\pi]$ with end point function value agreed. However, constant function is always inside. It suffices to check separating points property. However, it seems that separating $-\pi$ and $\pi$ value is not possible. So I do not see how to apply Stone-Wierstrass theorem to get contradiction. How to apply Stone-Weierstrass theorem here?
It is clear that I can convolve $e$ with fourier basis and get contradiction by fourier expansion convergence. 
Ref. Lang, Real and Functional Analysis Chpt IV, Sec 2, Banach Algebras.
 A: Assume $f$ could be an identity. Since the continuous functions are dense in $L^1([-\pi,\pi])$ (and are elements of $L^{\infty}([-\pi,\pi])$ by compactness), it would also be a unit for that convolution algebra. Define $g_n(x)=2\pi n 1_{[0,1/n]}.$ Then, $ g_n(x)=(g_n\star f)(x)= \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x-t) g(t)\textrm{d}t=n \int_0^{1/n} f(x-t)\textrm{d}t\to f(x)$ by continuity of $f$. Then, however, $f$ is $\infty$ at $0$ and $0$ else. There clearly is no such continuous function.
What's actually happening here is that $g_n(x)\to 2\pi \delta_{\{0\}}$ in the Weak-$^*$ Topology on $C([-\pi,\pi])$ which is the proper neutral element for the convolution algebra of finite measures.
A: Suppose there were such an identity $e.$ For $n\in \mathbb Z,$ let $f_n(t) = e^{int}.$ Then
$$1 = \hat {f_n}(n) = (e*f_n)\hat{\,}(n) = 
 \hat e(n)\cdot \hat {f}_n(n) = \hat e(n).$$
Thus $\hat e(n) =1$ for all $n.$ But $e\in L^1,$ so by the Riemann Lebesgue lemma, $\hat e(n) \to 0$ as $|n|\to \infty.$ This is a contradiction, proving there is no such $e.$
