$I=[a,b]$, let the function $f:I\rightarrow\mathbb{R}$ be convex.

(1) Is it possible to prove the existence of the limits: $$\lim_{x\rightarrow a^+}f(x) \ \ \ \ \ \lim_{x\rightarrow b^-}f(x)$$

If not, can we find a continuous function $g:I\rightarrow\mathbb{R} $ such that $g(x)=f(x)\ \forall x \in int(I)$?

I know that convexity implies that $f$ is continuous on the interior $int(I)$ of $I$ but I do not have any information on what happens at the boundary points $a$ and $b$.

(2) Moreover, how can I use the fact that the $left-derivative$ $f'_{-}$ of a convex function is $increasing$ in order to prove that each element of the following series is negative?

$$\sum_{j=0}^{y-1}\int_{\frac{2\pi j}{yk}}^{\frac{2\pi (j+1)}{ky}}f'_{-}(\frac{z}{k}) \frac{sin(z)}{k}dz$$


More detail than you ever wanted to know :-).

I am taking the interval to be $[0,1]$ to simplify some formulae. I am taking for granted the fact that a convex function is continuous on the relative interior of its domain. (This is straightforward to establish, but not the focus here.)

I need to establish two results first:

The first result shows that a function with a closed epigraph is lower semi continuous (lsc.). This is actually an equivalence, but we only need one direction here. Convexity is not used here.

Suppose $C \subset \mathbb{R}^{n+1}$ is closed and has the property that if $(x,t) \in C$ then $(x,s) \in C$ for all $s \ge t$ and we define $c:\mathbb{R}^n \to [-\infty, \infty]$ by $c(x) = \inf \{ t | (x,t) \in C \}$. Then $\operatorname{epi} c = C$. Furthermore, $c$ is lower semi continuous on $A= \{ x | (x,t) \in C \text{ for some } t \in \mathbb{R}^n \}$. Note that $A$ is not necessarily closed and $c(x) = +\infty$ for $x \notin A$..

To see this, suppose $(x,t) \in C$, then $c(x) \le t$ and so $(x,t) \in \operatorname{epi} c$. Now suppose $(x,t) \in \operatorname{epi} c$, then $\phi(x) \le t$. Hence there are $t_n \downarrow \phi(x)$ with $(x, t_n) \in C$. Since $C$ is closed, we see that $(x,\phi(x)) \in C$ and so $(x,t) \in C$.

Now suppose $x_n \in A$ and $x_n \to x$. Let $t_n = \phi(x_n)$ and $t = \liminf_n \phi(x_n)$. Suppose $t=+\infty$. Then we have trivially that $\phi(x) \le t$. Suppose $t \in \mathbb{R}$, then since $(x_n,t_n) \in C$, we see that $(x, t) \in C$, and so $\phi(x) \le t$. Finally, suppose $t=-\infty$. Then for any $M$ there is some subsequence such that $(x_n,M) \in C$ and so it follows that $(x,M) \in C$ and so $\phi(x) \le M$. Hence $\phi(x)=-\infty = t$.

The second result shows that under some mild conditions, a convex function is bounded below on a bounded, convex set.

Now suppose $A \subset \mathbb{R}^n$ is a bounded, non empty, convex set and $f:A \to [-\infty, \infty]$ is convex. In addition, suppose there is some $x \in \operatorname{ri} A$ such that $f(x) \in \mathbb{R}$. Then $f$ is bounded below.

There is no loss of generality in assuming $x \in A^\circ$. Since $f$ is convex, it is continuous on $A^\circ$ and there is some $\epsilon>0$ such that $\overline{B}(x,\epsilon) \subset A^\circ$. Since $\overline{B}(x,\epsilon)$ is compact, $\underline{f} = \min_{y \in \overline{B}(x,\epsilon)} f(y)$ is finite. Now let $y \in A \setminus \overline{B}(x,\epsilon)$, then $\underline{f} \le f(x+\epsilon {y-x \over \|y-x\|}) \le f(x) + \epsilon {1 \over \|y-x\|}(f(y)-f(x))$. Rearranging gives $f(y) \ge f(x) + {\|y-x\| \over \epsilon } ( \underline{f}- f(x))$, and since $A$ is bounded, we see that $f$ is bounded below.

Now suppose $f$ is as given in the question, then we see that $f$ is bounded below and if we let $E=\operatorname{epi} f$ and define $\phi(x) = \inf \{t | (x,t) \in \overline{E} \}$, then $\phi$ is also bounded below and is lsc. Since $\overline{E}$ is convex, it follows that $\phi$ is convex too. (The function $\phi$ is known in convex analysis as the closure of $f$.)

We see that $\phi(x) \le f(x)$ and if $x \in (0,1)$, then $\phi(x)=f(x)$. To see this, note that there is some $\epsilon>0$ such that $\overline{B}(x,\epsilon) \subset (0,1)$. Let $L=\overline{B}(x,\epsilon) \times \mathbb{R}$, then we have $L \cap \overline{E} = L \cap E$. One direction follows immediately, so suppose $(y,t) \in L \cap \overline{E}$. There are $(y_n,t_n) \in E$ such that $(y_n,t_n) \to (y,t)$ and hence by continuity of $f$, we have $f(y) \le t$, and so $(y,t) \in E$. Hence $\phi(x)=f(x)$.

If we can show that $\phi$ is also upper semi continuous (usc.) on $[0,1]$, then $\phi$ is continuous and the result follows.

To see this, note that if $x \in [{1 \over 2} , 1]$ then $x = (2x-1) + (1-(2x-1)) {1 \over 2}$ and so $\phi(x) \le (2x-1)\phi(1) + (1-(2x-1)) \phi({1 \over 2})$. Hence if $x_n \uparrow 1$, then $\limsup_n \phi(x_n) \le \phi(1)$ and so $\phi$ is lsc. at $x=1$. A similar analysis shows that the same holds true for $x=0$, and so $\phi$ is continuous. (This result can be generalised somewhat if the domain can be 'approximated' at the boundary by a suitable collection of simplices. See, for example, Theorem 10.4 in Rockafellar, "Convex Analysis".)

The above can be generalised slightly to allow $f$ to be $+\infty$ at $0,1$ (the limits may be $+\infty$, of course, for example, with $x \mapsto {1 \over x}$).

Simpler answer: Here is a proof that works for an interval on the real line. The function $f$ is not necessarily differentiable everywhere, but we do have that the slope (secant) is non decreasing (see here). Consider $x > { 1\over 2}$ and let $d(x) = {f(x)-f({1 \over 2}) \over x-{1 \over 2}}$, and note that $d$ is non decreasing, bounded above by $d(1)$ and hence we have the limit $d^* = \lim_{x \to 1} d(x)$. Since $f(x) = f({1 \over 2}) + d(x) (x- {1 \over 2})$, we see that $\lim_{x \to 1} f(x) = f({1 \over 2}) + {1 \over 2}d^*$.

The same sort of analysis applies to the other limit (or consider $x \mapsto f(1-x)$).

Original answer: (As Julian pointed out, this doesn't answer the question asked.) The function $f=1_{\{0,1\}}$ is convex but not continuous at the boundaries.

  • $\begingroup$ But $\lim_{x\to0^+}f(x)=\lim_{x\to1^-}=0$. $\endgroup$ Mar 16 '15 at 17:37
  • $\begingroup$ @JuliánAguirre: Thanks for catching that. Somehow I ended up in continuity land... $\endgroup$
    – copper.hat
    Mar 16 '15 at 17:44
  • $\begingroup$ @JuliánAguirre: I have repaired my answer. $\endgroup$
    – copper.hat
    Mar 16 '15 at 17:54
  • $\begingroup$ But I do think Filippo seems to be asking about what can happen at the boundaries. With your original answer you can demonstrate discontinuity. $\endgroup$ Mar 16 '15 at 18:04
  • $\begingroup$ @MichaelGrant: I added the original back... $\endgroup$
    – copper.hat
    Mar 16 '15 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.