{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Copy of s9

# Copy of s9 - Solutions to homework 9 1 By the uniform...

This preview shows page 1. Sign up to view the full content.

Solutions to homework # 9. 1. By the uniform continuity of f , there exists δ such that | f ( p ) - f ( q ) | < 1 whenever | p - q | < δ and p , q E . Since E is bounded, its closure E is both bounded and closed, hence compact, so the following open cover of E q E U q , where U q := { p : | p - q | < δ } has a finite subcover n k =1 U p k , which can also serve to cover E . Taking one point x k E from each U p k that contains points of E , we see that | f ( x ) | ≤ max k | f ( x k ) | + 1 for all x E . Thus f is bounded on E . The boundedness of E is indeed essential, as the example of the unformly continuous function f ( x ) := x on the domain E := IR demonstrates. 2. If f is continuous, then the function g : x f ( x ) - x is also continuous. If f (0) = 0, then f (0) > 0 and so g (0) > 0. If f (1) = 1, then f (1) < 1 hence g (1) < 0. Thus g takes values of opposite signs at the end of the interval I , hence g takes the value 0 somewhere inside I , whihc means f ( x ) = x for some x . 3. Both functions [ x ] and ( x ) are discontinuous at all integer points. Indeed, if k ZZ and x k - (i.e., x tends to k
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online