How to prove that the right derivative of a convex function is right continuous? let $f$ be a convex function and $D^+f$ be the right-derivative of $f$, I want to show that $D^+f$ is right continuous.
first I want to use the $\epsilon-\delta$: by the definition of $D^+f$, $\forall \epsilon,\exists h_0$ s.t. $\forall h<h_0$:
$$\frac{f(x+h)-f(x)}{h}-D^+f(x)<\epsilon\tag{1}$$
take $x_n\downarrow x$
$$ D^+f(x_n)-D^+f(x)=\lim_{h\to0^+}\frac{f(x_n+h)-f(x_n)}{h}-D^+f(x)\le \frac{f(x_n+h_0/2)-f(x_n)}{h_0/2}-D^+f(x)\tag{2}$$
I don't know how to find a relation between (1) and (2).
Can I take the limit of $n$ in (2) to get :
$$\lim_{n\to\infty}D^+f(x_n)-D^+f(x)\le \frac{f(x+h_0/2)-f(x)}{h_0/2}-D^+f(x)<\epsilon$$
then by the aribitrary of $\epsilon$?
 A: Say $f$ is convex on an open interval $I$ (and let's assume we've shown this implies that $f$ is continuous on $I$). Everything below refers to $f$ at (interior) points of $I$.$\newcommand\diff[2]{\frac{f(#2)-f(#1)}{#2-#1}}$.
Lemma 0. If $D^+f\ge0$ then $f$ is non-decreasing.
Proof: Suppose on the other hand that $a<b$ but $f(b)<f(a)$.
Convexity shows that  $$\diff ac\le\diff ab<0\quad(a<c<b).$$
Letting $c$ tend to $a$ from the right shows that $D^+f(a)<0$. QED.
Lemma 1. If $D^+f\ge\alpha$ and $a<b$ then $f(b)-f(a)\ge\alpha(b-a)$.
Proof: Apply Lemma 0 to $f(t)-\alpha t$. QED.
Theorem $D^+f$ is right continuous.
Proof: Convexity shows that $D^+f$ is non-decreasing. So it's enough to show that $$D^+f(a)\ge\lim_{b\to a^+}D^+f(b)=\alpha.$$ Choose $b>a$ with $b\in I$. Now, $D^+f\ge\alpha$ on $(a,b)$ (because $D^+f$ is non-decreasing), so Lemma 1 shows that $$f(y)-f(x)\ge\alpha(y-x),\quad a<x<y<b.$$Since $f$ is continuous this shows that $$\diff ay\ge\alpha,\quad(a<y<b).$$Now letting $y$ tend to $a$ from the right shows that $D^+f(a)\ge\alpha$. QED.
