Let $X$ be a locally convex topological vector space, $f, g : X \to \overline{\mathbb{R}}$ proper convex and lower semicontinuous.
Let $D \subset X$ be a (convex?) dense subset of $X$ such that $\left . f \right \vert_D = \left . g \right \vert_D$ and $f \left( x \right) \in \mathbb{R}$ for some $x \in D$.
Can we conclude that $f = g$?
In terms of epigraphs, it is clear that $\overline{\mathrm{epi} \left . f \right \vert_D} \subseteq \mathrm{epi} f \cap \mathrm{epi} g$, so that there is a largest proper convex and lower semicontinuous function $H: X \to \overline{\mathbb{R}}$ such that $\left . H \right \vert_D = \left . f \right \vert_D$.
Hence, we may conclude that $f \le H$ and $g \le H$, but I fail to see how one could deduce the reverse inequalities.
Note, that I do not want to assume that $f$ or $g$ is locally bounded nor locally finite.
Best Answer
No, this is not true. As an example take $X=L^2(0,1)$, \begin{align*} f(u)&=\begin{cases}\int_0^1|u'(x)|^2\,dx&\text{if }u\in H^1(0,1), u(0)=u(1)=0,\\\infty&\text{otherwise},\end{cases}\\ g(u)&=\begin{cases}\int_0^1|u'(x)|^2\,dx&\text{if }u\in H^1(0,1),\\\infty&\text{otherwise}.\end{cases} \end{align*} These are both proper lower semicontinuous quadratic (in particular convex) and coincide on the dense subspace $\{u\in H^1(0,1)\mid u(0)=u(1)=0\}$, where they are both finite. Still they do not coincide on all of $L^2(0,1)$.
Edit: Here is a proof of the lower semicontinuity of $g$. The proof for $f$ is similar, keeping in mind that $\{u\in H^1(0,1)\mid u(0)=u(1)=0\}$ is a closed subspace of $H^1(0,1)$.
Let $u_n\to u$ in $L^2$. We can assume that $(g(u_n))_n$ is bounded. Since $H^1$ is a Hilbert space, it follows that every subsequence of $(u_n)$ has a subsequence converging weakly in $H^1$. Since $u_n\to u$ in $L^2$, the only possible limit is $u$. Thus $u_n\to u$ weakly in $H^1$ and $\lVert u\rVert_{H^1}^2\leq \liminf_{n\to \infty}\lVert u_n\rVert_{H^1}^2$. Since $\lVert u_n\rVert_{L^2}\to \lVert u\rVert_{L^2}$, we can conclude $g(u)\leq \liminf_{n\to\infty}g(u_n)$.