Answer to Question #170337 in Real Analysis for Prathibha Rose

Question

Prove that every real valued convex function on the closed rectangle in R² is bounded

Answer 1

If x∈[a,b]

x∈[a,b] then we can write x=ta+(1−t)b

x=ta+(1−t)b with 0≤t≤1

0≤t≤1. Hence f(x)≤tf(a)+(1−t)f(b)≤t

max{f(a),f(b)}+(1−t)

max{f(a),f(b)}=

max{f(a),f(b)}

f(x)≤tf(a)+(1−t)f(b)≤tmax{f(a),f(b)}+(1−t)max{f(a),f(b)}=max{f(a),f(b)}

Proof of the fact that f

f is bounded below: On any closed sub interval of (a,b)

(a,b) the function is continuous, Hence bounded. Hence, if f

f is not bounded below then there exist a sequence xn

xn converging to

a or b such that f(xn)→−∞

f(xn)→−∞. Suppose xn→a

xn→a. Write (a+b)/2

 as λnxn+(1−λn)b.

Compute

λn from this equation and observe that λn→1/2

. Now take limits in

f(a+b)/2≤λnf(xn)+(1−λn)f(b) to get the contradiction

f(a+b)/2=−∞.