I want to show there's no identity for the convolution of two functions. I've found posts that answer this for Riemann integrable functions, but I'm working with the convolution given by $f\ast g = \int_{\mathbb{R}^d} f(x-y)g(y) dy$, where $f, g$ are Lebesgue integrable. So, I want to show there does not exist a $\delta\in{L^1(\mathbb{R}^d)}$ such that $f\ast\delta = f$.
My attempt:
Assume towards a contradiction that there is such a function $\delta$. Then we have $f\ast\delta = f$
and hence,
$$ \|\ f\ast\delta \|\ = \|\ f \|\ $$
but $$\|\ f \|\ \|\delta\|\ \ge \|\ f\ast\delta \|\ = \|\ f \|\ $$
which gives
$$ \|\ f\ast\delta \|\ = \|\ f \|\ $$
$$ \|\delta\|\ \ge \|\ f\ast\delta \|\ = 1 $$
With $ \|\ \delta \|\ \ge 1$, I believe I can derive a contradiction. I've tried integrating both sides, but that lead me nowhere. I'm not quite sure where to go from here. Any advice or tips?
Best Answer
Suppose there exists a function $\delta \in L^1_{loc}(\mathbb{R}^d)$ such that $\delta \ast f = f$ for all $f\in C_c^\infty(\mathbb{R}^d)$. Then let $\phi_k$ be a sequence of smooth functions with $0\leq \phi_k \leq 1$, $\phi_k(0)=1$ and each with support contained in $$ \left[-\frac{1}{k},\frac{1}{k}\right]. $$ The the dominated convergence theorem shows that $$ \int_{\mathbb{R}^d} \delta(x)\phi_k(x)\,\mathrm{d}x \to 0. $$ On the other hand, $$ 1 = \phi(0) = \delta\ast\phi(0) = \int_{\mathbb{R}^d} \delta(x)\phi_k(x)\,\mathrm{d}x $$ which is clearly a contradiction.
The above gives a slightly stronger result then what you asked for.
Alternatively, suppose for the sake of contradiction that there exists a function $\delta \in L^1(\mathbb{R}^d)$ such that $f\ast \delta = f$ for all $f\in L^1(\mathbb{R}^d)$.
Then, consider the simple $L^1$-function $\rho = \mathbb{1}_{S}$ where $$ S= \left[-\frac{1}{2},\frac{1}{2}\right]^d $$ For almost every $x\in S$, we must have $$ 1 = \rho\ast\delta(x) = \int_{\mathbb{R}^d}\rho(x-y) \delta(y)\,\mathrm{d}y =\int_{S+\{x\}} \delta(y)\,\mathrm{d}y. $$ An application of the dominated convergence theorem therefore shows that $$ \int_{S} \delta(y)\,\mathrm{d}y = 1. $$ Similarly, for every $x_0\in \mathbb{Z}^d$ we can derive $$ \int_{S_{x_0}} \delta(y)\,\mathrm{d}y = 1 $$ where $$ S_{x_0}= \left[x_0-\frac{1}{2},x_0+\frac{1}{2}\right]^d. $$ Finally, we conclude that $$ \int_{\mathbb{R}^d} \delta(y)\,\mathrm{d}y =\sum_{x_0\in\mathbb{Z}^d}\int_{S_{x_0}} \delta(y)\,\mathrm{d}y = \infty $$ which is clearly a contradiction.