# 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

\label{Theorem34eq}

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

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}