Epigraph of a function. I hope you can give me some suggestions on convex functions.
the function $f:(0,\infty)\rightarrow \mathbb{R}$ given by $f(x)=\dfrac{1}{x}$ is convex and continuous, but its epigraph is closed in $\mathbb{R}^{2}$ ?. 
 A: Let $\phi(x,y) = xy$. Then $\phi$ is continuous and $\phi^{-1} [1,\infty)$ is closed. Hence $\text{epi} f = \phi^{-1} [1,\infty) \cap [0,\infty)^2$ is closed.
A: In general, if $U\subset \mathbb R$ and $f:U\rightarrow \mathbb R$, the the epigraph is the set $\{(x,a):x\in U, a>=f(u)\}$.
What conditions on $f$ and $U$ are needed to ensure that this set is closed in $\mathbb R^2$?
In this case, the reason the epigraph is closed is that near the boundary of $U$, $f$ approaches infinity. If there is a point on the boundary of $U$ which is not in $U$ where $f$ does not appoach $+\infty$, then the epigraph would not be closed. It is precisely because $U$ has one boundary point, $0$, and $lim_{x\to 0} f(x) = +\infty$ that this epigraph is closed.
The epigraph is a subset of $U\times\mathbb R$, so the closure in $\mathbb R^2$ is necessarily a subset of $\bar U\times\mathbb R$, where $\bar U$ is the closure of $U$. The epigraph is closed in $U\times\mathbb R$, so the only points in this closure that cannot be in the epigraph are points of the form $(x,y)$ where $x\in\bar U\setminus U$.
If there is such an $(x,y)$, then there has to be $(x_i,a_i)$ in the epigraph such that $x_i\to x$ and $a_i\to y$. But then it is impossible for $\lim_{x'\to x} f(x')=+\infty$.
On the other hand, if it is not true that $\lim_{x'\to x} f(x')=+\infty$, then for some $N$, for all $n>0$ there is an $x_n$ such that $|x-x_n|<\frac{1}{n}$ with $f(x_n)\leq N$.
But then $(x_n,N)$ is in the epigraph, so $(x,N)$ would be in the closure, so the epigraph would not be closed.
So the epigraph of $f$ is closed in $\mathbb R^2$ iff the epigraph is closed in $U\times\mathbb R$ and the limits towards all the boundary points of $U$ is $+\infty$.
A: We can also use the fact that a function $g:\mathbb R^n \to [-\infty,\infty]$ is closed if and only if all of its sublevel sets are closed.
If $\alpha > 0$ then 
\begin{align*}
\{ x \in \textbf{dom } f\mid f(x) \leq \alpha \} &= \{x \in \mathbb R| \frac{1}{x} \leq \alpha \text{ and } x > 0 \} \\
&= \{ x \in \mathbb R \mid \frac{1}{\alpha} \leq x \}
\end{align*}
which is closed.
Moreover, if $\alpha \leq 0$, then the sublevel set $\{ f \leq \alpha \}$ is empty.
This shows that all the sublevel sets of $f$ are closed, so $f$ is closed.
