Proving Continuity and Intermediate Value Theorem in Real Analysis

[squaremeplz]

Homework Statement

1. f is a continuous real valued function on a closed interval. f assumes minimum fvalues on [a,b], there exit x_o,y_o in [a,b] such that f(x_o) <= f(x) <=f(y_o) for all x in [a,b]. Show that the function -f assumes its maximum at x_o in [a,b], then f assumes its minimum at x_o.

2. Suppose that f is continuous on [0,2] and that f(0) = f(2). Prove that there exist x,y in [0,2] such that |x-y| = 1 and f(x) = f(y). Hint: consider g(x) = f(x+1) - f(x) on [0,1]

Homework Equations

The Attempt at a Solution

1. clearly since f(y_o) <= -f(x) <=f(x_o)

multiplying this inequality through by - we get

-f(y_o) <= f(x) <=-f(x_o)

or

f(x_o) <= f(x) <=f(y_o) for all x in [a,b]

2. Hint: consider g(x) = f(x+1) - f(x) on [0,1]

since:

g(0) = f(1) - f(0)
g(1) = f(2) - f(1)

since: f(2) = f(0)

g(0) = f(1) - f(0)

g(1) = f(0) - f(1)

this shows that

g(0) + g(1) = 0

or

g(0) = -g(1)

then

f(1) - f(0) = -[f(0) - f(1)]

f(1) - f(0) = f(1) - f(0)



therefore f(1) = f(0) and 0,1 are in [0,2] and satisfy |y-x| = 1

is this anywhere close? any help would be greatly appreciated for I spent much time trying to figure out this one. Thanks again!

 

[Pere Callahan]

therefore f(1) = f(0) and 0,1 are in [0,2] and satisfy |y-x| = 1

This conclusion doesn't make any sense.

g(0)=-g(1)

Take it from here.
By the intermediate value theorem, you know that there exists $x_0\in[0,1]$ with $g(x_0)=0$.
How can you go on from here?