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Proof: \(\label{freeEnergyEquation} f'(x) = \lambda(x).\)

\begin{proof}

Let \(f_N\) denote the free energy for the gradient model on \)\Lambda_N\),

\[

f_N(x) = \lim_{N \rightarrow \infty} \frac{1}{\beta |\Lambda_N| } \log Z_N^x.

\]

In \cite{FS_97}, Lemma 3.1 (ii) gives the identity

\[

f_{N}'(x)=\mathbb{E}_{\mu_{N}^x}[V'(\eta+x)],

\]

(whilst this is proven in the context of conditions \eqref{Cond5}-\eqref{Cond4}, it does not require the convexity assumption or more regularity than \(V\in C^{1}\), therefore it holds in our case). Further, Theorem 3.4 (iii) of the same reference gives us

\begin{equation}\label{Theorem34eq}

f'(x)=\mathbb{E}_{\mu^{x}}[V'(\eta+x)]

\end{equation}

for the limit measure \(\mu^x\). Since the expectation on the right-hand side of \eqref{Theorem34eq} is for a single bond, then using the factorisation obtained in Lemma \ref{productCramer} we obtain

\[

f'(x)=\mathbb{E}_{\hat{\nu}_{\lambda}}[V'(\eta+x)].

\]

where \(\lambda = \lambda(x)\). Finally, we use integration by parts to evaluate the expectation

\begin{align*}

f'(x) &= \frac{1}{\widehat{Z}_{\lambda}} \int_{\mathbb{R}} V'(\eta+x) e^{-V(\eta+x)} e^{\lambda\eta} d\eta \\

&= \left. -\frac{e^{\lambda \eta -V'(\eta+x)}}{\widehat{Z}_{\lambda}}\right|_{-\infty}^{\infty} + \frac{\lambda}{\widehat{Z}_{\lambda}} \int_{\mathbb{R}}e^{\lambda \eta -V'(\eta+x)} d\eta \\

&= \lambda,

\end{align*}

where we require \(\eqref{Cond1}\) to ensure that the first term vanishes, as we saw in Remark \ref{vanishingRemark}.

\end{proof}