Problem #23 pg-94, Stein and Shakarchi As an application of the Fourier transform, show that
there does not exist a function $I\in L^1(R^d,m)$ such that
$f*I = f$ for all $f\in L^1(R^d,m)$.
 A: If you know Fourier transform, and in particular how it behave with respect to convolution product, the question is quite easy. Indeed, if $g$ and $h$ are integrable functions, then the relationship $\widehat{f*g}(x)=\widehat f(x)\widehat g(x)$, and the Fourier transform of an integrable function is continuous .
Otherwise, you can consider mollifiers, that is, non-negative smooth function with support contained in $B(0,n^{-1})$, of integral $1$. Take $(\phi_n)$ such a sequence, and show that $\phi_n*I\to I$ converge almost everywhere to $I$. 
A: If $ I \in L^1 $ then from uniform integrability given any $\epsilon >0 $ there is a $ \delta(\epsilon) > 0 $ such that you have $ \int_E |I(x)|dm <\epsilon $ for all measurable $E$ with $ m(E) < \delta(\epsilon) $. Now choose characteristic function $ f_\delta = \chi_{B(0,\delta)}\in L^1 $ and assume your statement is true for some $ I \in L^1 $. Then you have to have $$\int_{\mathbb{R}^d} f_\delta(x-y)I(y)dy = f_\delta (x) $$ Take $ x = 0 $ in the above and choose $ \delta$ such that $ m(B(0,\delta))= \delta(\epsilon)/2 $ you have the contracdiction
$$ 1 = f_\delta(0) = \int_{\mathbb{R}^d} f_\delta(-y)I(y)dy = \int_{B(0,\delta)}I(y)dy <\epsilon $$
