Prove that $L^1([-\pi, \pi))$ (the space of integrable $2\pi$ periodic functions using the norm $\lVert f\rVert_1 = \frac{1}{2\pi}\int_{-\pi}^\pi |f(x)|dx$) is not unital under the convolution product. That is, there does not exist $k\in L^1([-\pi,\pi))$ so that $k\star f = f \star k = f$ for all $f \in L^1([-\pi,\pi))$, where $f\star g (\theta):= \frac{1}{2\pi} \int_{-\pi}^\pi f(\theta – t) g(t)dt$ denotes the convolution product.
I think an approach is a proof by contradiction. It might be useful to use the fact that $k\star f (0) =f\star k(0) = f(0)$ to deduce some useful properties about $k$ that can lead to a contradiction. By definition, $k\star f (0) = \frac{1}{2\pi} \int_{-\pi}^\pi k(-t) f(t)dt$. If we then choose $f$ to be continuous, then I think one can show that $k(t)$ must vanish out of some set $(-\delta, \delta), \delta < \pi$. In fact, one might be able to show that since the claim holds for all continuous functions, then $k$ must vanish everywhere except possibly at $0$. But how would one justify this more formally (e.g. maybe something like the uniform convergence of continuous functions on bounded intervals might be useful)?
Best Answer
Suppose that such a $k \in L^1(-\pi, \pi)$ exists.
Then apply the identity to the Fourier exponentials: For all $n\in \mathbb Z$, $$\frac 1{2\pi} \int_{-\pi}^\pi k(t)e^{in(\theta-t)}dt=e^{in\theta} $$ This simplifies as $$\frac 1{2\pi} \int_{-\pi}^\pi k(t)e^{-int}dt=1 $$ In other words, the Fourier coefficients of $k$ are all $1$.
But that's not possible, because the Riemann-Lebesgue lemma tells us that the Fourier coefficients of an $L^1(-\pi, \pi)$ function converge to $0$ as $|n|\rightarrow +\infty$.