L^{\ast} \leq  L^{\ast} L + L L^{\ast} \ . ~~~~~~~~~~~~~~~\blacksquare
\label{eq:2.10}$$

Returning to the proof of Theorem \[theo:nonrel\] we start with the lower bound. Denote by $$\Pi(x) = \frac{-i}{\sqrt{2V}}\sum_{|k| < \Lambda , \lambda}
{\sqrt{|k|} \varepsilon_{\lambda}(k) } \left( a_{\lambda}(k) 
e^{ik \cdot x} - a^{*}_{\lambda}(k) e^{-i k \cdot x} \right) \ .
\label{eq:2.11}$$ This operator is canonically conjugate to $A(x)$ in the sense that we have the commutation relations $$i[\Pi_i(x) , A_j(x)] = \delta_{i,j} \frac{1}{(3\pi)^2 } \Lambda ^3 \ .
\label{eq:2.12}$$ For our calculation, it is important to note the $${\rm div} ~ \Pi(x) =0 \ .
\label{eq:2.13}$$ Hence from (\[eq:2.12\]) and (\[eq:2.13\]) we get that (for all $j$) $$[p_j + \sqrt{\alpha} A_j(x) , \Pi_j(x)] = \sqrt{\alpha} \frac{i}{(3\pi)^2 } \Lambda ^3 \ . 
\label{eq:2.14}$$ The inequality $$\frac{1}{2 } (p+ \sqrt{\alpha}A(x))^2 + 2a^2 \Pi(x)^2 \geq
-ai \sum^{3}_{j=1}[p_j + \sqrt{\alpha} A_j(x) , \Pi_j(x)] \ ,
\label{eq:2.15}$$ valid for all positive numbers $a$, yields $$\frac{1}{2} (p+ \sqrt{\alpha} A(x))^2 + H_f 
\geq a \sqrt{\alpha} \frac{1}{(3 \pi)^2} \Lambda ^3 + H_f - 2a^2 \Pi(x)^2 \ .
\label{eq:2.16}$$ Now, with $$v_ {\lambda} (k) = (3 \pi)\frac{\varepsilon_{\lambda}(k)}{\Lambda ^{3/2}}
\label{eq:2.17}$$ and $$w(y) = \delta (x-y) \ ,
\label{eq:2.18}$$ it is elementary to see that (\[eq:2.7\]) is satisfied. Hence Corollary \[corollary:1\] 