Let $G$ be a locally compact group (not discrete) and let $L$ be the left regular representation of $A = L^1(G)$ on itself i.e. $L: A \to \mathcal{B}(A)$ where $L(f): A \to A$, $L(f)(g) = f*g$. I want to show that $\forall f\in A$, $||L(f) - I|| \geq \frac{1}{2}$ where $I$ is the identity operator on A.
Using the fact that $L^1(G)$ has an approximate identity, one can show that $||L(f) - I|| \geq | ||f||_1 - 1|$ and so the problem is reduced to $f \in A$ such that $\frac{1}{2}< ||f||_1 < \frac{3}{2}$. I'm not entirely sure how useful this is, but it's the only thing I've been able to come up with so far. Any hints or pointers in the right direction would be much appreciated.
Edit: I'm still quite lost on this problem. I've tried simply considering the case where $G = \mathbb{R}$. In this case I've managed to show that it's true if $f$ is an indicator function on an interval, but the method I've been using falls apart when considering finite linear combinations of such functions.