Solve ivt for non-continuous functions

[orangesun]

Homework Statement

We take f:[0,1]→[0,1] a non-increasing function, such that f(x)≥f(y) whenever x≤y; and we want to prove that there exists c∈[0,1] such that f(c)+c=1. We letA={x∈[0;1]:f(x)+x≥1} and we define c=infA.

a) explain why c exists

b)let xn be a sequence of elements of A that converges to c, prove that for all n,f(c)+xn≥1
c)deduce that f(c)+c≥1
d)if c=0 prove that f(c)+c≤1
e)if c>0, consider xn=c−1/n prove that f(c)+c≤1

Homework Equations

The Attempt at a Solution

Hi, I have been having trouble solving this question since the only information i could find online was using the IVT properties. Apparently this question should be solved without using the properties.
for
a) i have said that since the set A is bounded, and there is a lower bound, then there must be an inf for A and hence c must exsist
b) for b i am not quite sure, as i know that xn can be written in terms of c

So far i have solved something by the lines of:

since f(x) + x [0,1] and we are given to find f(x) + x $\geq$ 1,
f(0) + 0 $\geq$ 1;
f(1) + 1 $\geq$ 1;
hence f(1) $\leq$ c $\leq$ f(0)

I just need help and some direction on where to go next. I am really stuck. Thank you in advance!

 

[mighty2000]

Hi there, let me try to help you with this problem.

a) You are correct in saying that c exists because A is bounded and has a lower bound. In fact, A is a closed and bounded subset of [0,1], so c must exist by the completeness property of the real numbers.

b) To prove that f(c) + xn ≥ 1 for all n, let's start by looking at the definition of c. We know that c = infA, which means that for any ε > 0, there exists an element xε in A such that c ≤ xε < c + ε. Now, since xn is a sequence in A that converges to c, we know that for any ε > 0, there exists an index N such that for all n ≥ N, xn ∈ [c, c+ε). In other words, xn is always "close" to c. Now, using the non-increasing property of f, we can say that for all n ≥ N, f(c) ≥ f(xn). Combining this with the fact that xn ≥ c, we get f(c) + xn ≥ f(xn) + xn ≥ f(c) + c ≥ 1, where the last inequality follows from the definition of A. This holds for all n ≥ N, so it holds for the entire sequence.

c) From part b, we know that f(c) + xn ≥ 1 for all n. Taking the limit as n approaches infinity, we get f(c) + c ≥ 1, since xn converges to c.

d) If c = 0, then by definition, f(c) + c = f(0) + 0 = 1. So, f(c) + c ≤ 1.

e) If c > 0, then consider the sequence xn = c - 1/n. This sequence is in A, since for all n, f(xn) + xn = f(c - 1/n) + c - 1/n ≥ 1, where the inequality follows from the definition of A. Now, using the same argument as in part b, we can show that f(c) + xn ≥ 1 for all n, and taking the limit as n approaches infinity, we get f(c) + c ≤ 1.

Hope this helps!