Proof of continuity of convex functions

[shotputer]

Homework Statement

Let a function f : R => R be convex. Show that f is necessarily continuous. Hence, there can be no convex functions that are not also continuous.

Homework Equations

The Attempt at a Solution

F is continuos if there exist $$\epsilon$$ >0 and $$\delta$$>0 such that |x-y|< $$\delta$$ => |f(x) - f(y)| <$$\epsilon$$

on the other hand function is convex if for some numbers a and b, with a<b an $$\lambda$$ $$E$$ (0,1) f( $$\lambda$$ a+ (1-$$\lambda$$ b) $$\leq$$ $$\lambda$$ f(a) + (1-$$\lambda$$) f(b)

so basicaly that is what I have, I got hint that we should use this definition of continuity... but I don't have idea where to start...
help, pleease...

 

[ftdhk]

The statement does not work if f is defined on [a, c] since

f(x) = 0 for x in [0,1)
f(1) = 1 for x=1

is a valid convex function (and continuous on (0,1)). A strictly convex function can be generated by a strictly convex function on [0,1) with f(1+) < 1. It does work for f on (a,c).

I can give you two directions, but it still needs a lot of work.

1) This is something I found on the web. Directly, let f be a convex function on (a,c) with a<b<c. The draw a line m1 through a and b, and another line m2 through a and c. We know f((a,b)) < m1 and f((b,c)) < m2. Show that the convex function lies between the two lines (m2 < f((a,b)) and f((b,c) < m1). I have yet to prove this. Then the intersection is be, and the function approaches b from both sides. Hence it is continuous at b.

Like I said, proving the function lines between both lines is something I have not done, but worth a try.

2. I have a professor who uses the contrapositive, but we need a couple of assumptions, and it still is not easy.

If f is a convex function on (a,c) then it must be bounded below (a or c can be + inf, think csc x on (-pi/2 pi/2). The proof is hard, but a sketch is easy to see ( Try drawing a convex function with a or c going to -inf).

Next, we have to show that convex function with a minimum on a bounded interval (csc), has at most, a countable number of discontinuities with intervals. (Could we construct a function defined on (a,c) but discontinuous on all (a,c)). I have not figured that out. If someone can do this, then the statement is not true. Now there are pathelogical functions that are continuous on an uncountable set (irrational numbers) and discontinuous on the rational set. If we can show that the construct is continuous on an interval, then the proof becomes easier "to see".

If a function f on (a,c) is discontinuous at b, and there exist intervals (b-d1,b) and (b,b+d2) such that the function is continuous, then f is not convex.

f must be convex for (b-d1,b) and (b,b+d2). However, from (b-d1 to b+d2) there will be points above the line. This needs to be proven.



ftdhk.

 

[zzzhhh]

Here is a proof that convex function is necessarily continuous. This proof is actually a complement of the first direction of ftdhk' post. As s/he said, we need only have to prove that the function lies between the two lines. As indicated by ftdhk, we prove it on an open interval $I$. Assume $a<b<c$, we want to prove $\frac{f(b) - f(a)}{b - a} \le \frac{f(c) - f(a)}{c - a} \le \frac{f(c) - f(b)}{c - b}$. Let $\lambda=\frac{c - b}{c - a}$, then $1-\lambda=\frac{b - a}{c - a}$ and $b=\lambda a+(1-\lambda)c$. By the definition of convex function, $f(b)\leq \lambda f(a)+(1-\lambda)f(c)=\frac{c - b}{c - a}f(a)+\frac{b - a}{c - a}f(c)$. Multiplying $c-a$ (>0) on both sides, we get $(c-b)f(a)+(b-a)f(c)-(c-a)f(b)\geq0$. Express this in determinant form, we have $\left| {\begin{array}{*{20}{c}} 1 & a & {f(a)} \\ 1 & b & {f(b)} \\ 1 & c & {f(c)} \\ \end{array}} \right|\geq0$. But $\left| {\begin{array}{*{20}{c}} 1 & a & {f(a)} \\ 1 & b & {f(b)} \\ 1 & c & {f(c)} \\ \end{array}} \right|=\left| {\begin{array}{*{20}{c}} 1 & a & {f(a)} \\ 0 & {b - a} & {f(b) - f(a)} \\ 0 & {c - a} & {f(c) - f(a)} \\ \end{array}} \right|=(b-a)(f(c)-f(a))-(c-a)(f(b)-f(a))\geq0$. This is just $\frac{f(b) - f(a)}{b - a} \le \frac{f(c) - f(a)}{c - a}$. Similarly, $\left| {\begin{array}{*{20}{c}} 1 & a & {f(a)} \\ 1 & b & {f(b)} \\ 1 & c & {f(c)} \\ \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 0 & {a - c} & {f(a) - f(c)} \\ 0 & {b - c} & {f(b) - f(c)} \\ 1 & c & {f(c)} \\ \end{array}} \right|=(a-c)(f(b)-f(c))-(b-c)(f(a)-f(c))\geq0$, This is equivalent to $\frac{f(c) - f(a)}{c - a} \le \frac{f(c) - f(b)}{c - b}$. Then we obtain the desired inequalities. From this inequalities, we can prove that $f(x)$ stands above the line connecting $(a,f(a))$ and $(b,f(b))$ for all x outside (a,b): suppose c>b and the value on the line of c is $l(c)$, then $\frac{f(b) - f(a)}{b - a} = \frac{l(c) - f(b)}{c - b}$. If $f(c)<l(c)$, then $\frac{f(c) - f(b)}{c - b}< \frac{l(c) - f(b)}{c - b}=\frac{f(b) - f(a)}{b - a}$, which violates the inequality just proved. The same argument applies for c<a. This justifies the assertion that the function f lies between two piecewise linear lines, and $\lim \limits_{x \to b} f(x) = f(b)$ by the so-called "squeeze theorem", which is just the fact that f is continuous at b, and then on the whole $I$ since b is an arbitrary element of $I$.