Left regular representation of $L^1(G)$ for a locally compact group $G$ 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.
 A: You seem happy that $L^1(G)$ has a bounded approximate identity.  Let me be a little more precise-- there is a net $(e_\alpha)$ of norm one elements of $L^1(G)$ such that $ae_\alpha \rightarrow a$ and $e_\alpha a\rightarrow a$ in norm, for each $a\in L^1(G)$.  So for $f\in L^1(G)$,
$$ \|L(f)-I\| \geq \liminf_\alpha \|L(f)e_\alpha - e_\alpha\| = \liminf_\alpha \|f-e_\alpha\|. $$
Now I'd actually use some construction of $e_\alpha$.  If $G=\mathbb R$, then then we can choose a sequential bai, say
$$ e_n = 2n\chi_{[-1/n,1/n]}.$$
Then
$$ \|f-e_n\|_1 = \|f\chi_{(-\infty,-1/n)}\| + \|f\chi_{(1/n,\infty)}\|
+ \int_{-1/n}^{1/n} |f-2n| $$
By monotone convergence the first two terms converge to $\|f\|$.  Use the reverse triangle inequality on the final term to bound it from below by
$$ 1 - \int_{-1/n}^{1/n} |f| $$
which converges to $1$ by dominated convergence.  So
$$ \|L(f)-I\| \geq \|f\|+1. $$
For a general $G$, just choose the obvious generalisation of this construction of a bai (thanks to t.b. for suggesting this change of wording!)
Of course, if you only need $1/2$ then there might be an easier argument...
Edit:  I'm tempted to use that $\|L(f)-I\|<1/2$ implies that $L(f)$ is invertible, but I don't quite see how to use this right now...
