# Solve ivt for non-continuous functions

1. Apr 22, 2013

### orangesun

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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!

Last edited: Apr 22, 2013