Theorem 7.5 from Rockafellar I am trying to understand the proof from Theorem 7.5 from Rockafellar's "Convex Analysis". The Statement is the following:
Let $f$ be a proper convex function and let $x\in\text{ri(dom(}f))$. Then:
$$\text{cl}f(y)=\lim_{\lambda\uparrow 1}f((1-\lambda)x+\lambda y)$$
for every $y$.
While showing that $$\text{cl}f(y)\geq\limsup_{\lambda\uparrow1}f((1-\lambda)x) $$
he defines $\beta\geq\text{cl}(f)(y)$. Then, I understood that in the end of the proof he concludes that 
$$\limsup_{\lambda\uparrow1}f((1-\lambda)x)\leq \beta.$$
And then concludes the proof.
Alternative proofs are also welcome!
Thanks!
 A: The basic idea is to show that the limit exists by breaking it into $\lim\inf$ and $\lim\sup$. We ultimately want to have the following,
$$\mbox{cl}f(y)\leq\lim\limits_{\lambda\uparrow 1}\inf f((1-\lambda)x+\lambda y)\leq\lim\limits_{\lambda\uparrow 1}\sup f((1-\lambda)x+\lambda y)\leq \mbox{cl}f(y)$$
Now, the left half comes for free since $f$ is a majorant of $\mbox{cl}f(y)$,
\begin{align}
\mbox{cl}f((1-\lambda)x+\lambda y)&\leq f((1-\lambda)x+\lambda y)\\
\implies \lim\limits_{\lambda\uparrow 1}\inf \mbox{cl}f((1-\lambda)x+\lambda y)&\leq \lim\limits_{\lambda\uparrow 1}\inf f((1-\lambda)x+\lambda y)
\end{align}
and since $\mbox{cl}f(y)$ is a lower semicontinuous function (by definition),
$$\mbox{cl}f(y)\leq\lim\limits_{\lambda\uparrow 1}\inf \mbox{cl}f((1-\lambda)x+\lambda y)\leq\lim\limits_{\lambda\uparrow 1}\inf f((1-\lambda)x+\lambda y)$$
So we only have to show the right half of our inequality chain and we are done. Let $\beta$ be a real number such that $\beta \geq \mbox{cl}f(y)$ (remember $y$ is fixed), then $(y,\beta)$ must be in the epigraph of $\mbox{cl}f$ by definition, $(y,\beta)\in\mbox{epi cl}f$. We have from a previous lemma that we can swap the order of cl and epi for suitably behaved functions (i.e. $f$),
$$(y,\beta)\in\mbox{cl epi}f$$
This is an important step because now we are working entirely in the epigraph of just $f$ instead of both $f$ and cl$f$. 
We are going to connect two points in the epigraph of $f$, we already have one $(y,\beta)$ so let $\alpha>f(x)$ to give us another one, $(x,\alpha)\in\mbox{epi}f$. Since $x\in \mbox{ri dom}f$ we have that $(x,\alpha)\in\mbox{ri epi}f$ and therefore $(1-\lambda)(x,\alpha)+\lambda (y,\beta)\in\mbox{ri epi}f$ for all $0\leq\lambda<1$. This inclusion tells us that the line segment connecting $(x,\alpha)$ and $(y,\beta)$ is in the relative interior of the epigraph of $f$ except for possibly the end point $(y,\beta)$. Since the rest of the line segment is contained in the relative interior of the epigraph of $f$, we are free to use the following inequality,
$$f((1-\lambda)x+\lambda y)\leq (1-\lambda)\alpha + \lambda\beta\quad 0\leq\lambda<1$$
as the epigraph of $f$ is a convex set. All that is left to do now is take the lim sup over the inequality (which holds for all the values of $\lambda$ we are considering in the lim sup),
$$\lim\limits_{\lambda\uparrow 1}\sup f((1-\lambda)x+\lambda y)\leq \lim\limits_{\lambda\uparrow 1}\sup(1-\lambda)\alpha+ \lambda\beta=\beta$$
This holds for all $\beta \geq \mbox{cl}f(y)$ so in particular we can take $\beta=\mbox{cl}f(y)$ and the proof is complete. It just now dawned on me this is probably the only line you were missing.
