hw1 - Problem 5: Let f;g : R ! R be continuous functions....

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

View Full Document Right Arrow Icon
Math 125A, Fall 2009 Homework 1 Due Date: October 2, 2009 Problem 1: Prove, using either De±nition 17.1 or Theorem 17.2, that each function is con- tinuous at x = c . (a) f ( x ) = x 2 and c = 2 (b) f ( x ) = p x and c = 0 (c) f ( x ) = ( x sin( 1 x ) ; if x 6 = 0 0 ; otherwise and c = 0 Problem 2: Prove, using either De±nition 17.1 or Theorem 17.2, that each function is not continuous at x = c . (a) f ( x ) = ( sin( 1 x ) ; if x 6 = 0 0 ; if x = 0 and c = 0 (b) f ( x ) = b x c and c 2 Z (Here b x c is the largest integer n satisfying n ± x .) Problem 3: Let f;g : R ! R be continuous functions. (a) Prove that min( f;g ) = 1 2 ( f + g ) ² 1 2 j f ² g j . (b) Prove that h = min( f;g ) is a continuous function. Problem 4: De±ne functions f : R ! R and g : R ! R as follows: f ( x ) = ( 1 ; if x 2 Q 0 ; otherwise and g ( x ) = ( x; if x 2 Q 0 ; otherwise (a) Prove that f is discontinuous everywhere. (b) Prove that g is continuous at x = 0 but discontinuous everywhere else.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 5: Let f;g : R ! R be continuous functions. (a) Prove that if f ( r ) = 0 for all r 2 Q , then f ( x ) = 0 for all x 2 R . (Hint: Proof by contradiction.) (b) Prove that if f ( r ) = g ( r ) for all r 2 Q , then f ( x ) = g ( x ) for all x 2 R . Problem 6: Let f : R ! R be an additive function, i.e., f satises the functional equation: f ( x + y ) = f ( x ) + f ( y ) for any x;y 2 R . Furthermore, suppose that f has at least one point of continuity c 2 R . Prove that, in fact, f is continuous on the entire real line....
View Full Document

Ask a homework question - tutors are online