\sqrt{|k|}
a_{\lambda}(k) \overline{v}_{\lambda} (k)e^{i k \cdot y} \ ,
\label{eq:2.4}$$ with some arbitrary complex coefficients $v_{\lambda}(k)$. The following lemma is elementary

$$H_f \geq \int w(y) L^{*}(y) L(y) {\rm d} y\ 
\label{eq:2.5}$$

provided that the functions $v_{\lambda}(k)$ and $w$ are chosen such that, as matrices, $$|k|\delta_{\lambda^{\prime}, \lambda} \delta(k^{\prime}, k)
\geq  \frac{1}{2V} \overline{v}_{\lambda} (k) 
\widehat{w}(k-k^{\prime}) v_{\lambda^{'}} (k^{'}) \ ,
\label{eq:2.6}$$ or equivalently, that $$\sum_{|k| < \Lambda, \lambda} \frac{|f_{\lambda}(k)|^2}{|v_ {\lambda} (k)|^2}
 \geq \sum_{|k|,|k^{'}| < \Lambda, \lambda, \lambda ^{'}} \frac{1}{2V}
\overline{f_{\lambda}(k)}f_{\lambda^{\prime}} (k^{\prime})
\widehat{w}(k-k^{\prime})
\label{eq:2.7}$$ for all $f_{\lambda}(k)$, where $\widehat{w}(k) = \int e^{ik \cdot x} w(x) {\rm d} x$. \[lemma:2.2\]

For the proof, one simply notes that both sides of (\[eq:2.5\]) are quadratic forms in the creation and annihilation operators, and hence (\[eq:2.6\]) and (\[eq:2.7\]) are necessary and sufficient conditions for (\[eq:2.5\]) to be true.

$$H_f \geq -\frac{1}{2V} \sum_{|k| < \Lambda, \lambda} |k|
|v_{\lambda} (k)|^2 \ \int w(y) {\rm d} y + \frac{1}{4}
\begin{cases}
\int w(y) (L(y) + L^{*}(y))^2 {\rm d} y \\
 -\int w(y) (L(y) - \frac{1}{4} L^{*}(y))^2 {\rm d} y 
\end{cases}
\label{eq:2.8}$$

\[corollary:1\]

To prove this, note that $$L^{*} L = L L^{*}- \frac{1}{2V} \sum_{|k| < \Lambda , \lambda} 
|k| |v_{\lambda} (k)|^2 \ ,
\label{eq:2.9}$$ and, quite generally for operators, $$LL + L^{\ast} 