This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Mathematics 218: Analysis in Several Variables Fall Term 2009 First Problem Set Rik Sengupta September 30, 2009 1 Group 1 1. Sketch the following subsets of R 2 = { x = ( x 1 ,x 2 ) } ; and for each subset indicate whether it is open, closed, or compact, and find its boundary. (a) { x  < k x k 1 } . (b) { x  < x 1 1 ,x 2 = sin 1 x 1 } . (c) { x k x k 1 } { x  < k x k 2 < 1 } . (d) { x  x 1  1 } { x  x 2 1 } . Solution. 1 [contd] 2 2. Indicate the points at which each of the following functions in R 2 is con tinuous. (a) f ( x ) = x 2 sin x 1 if x 1 x 1 + x 2 if x 1 < (b) f ( x ) = x 1 + x 2 if k x k 1 x 2 1 + x 2 2 if k x k > 1 Solution. (a) This function is continuous everywhere on R 2 . (b) Consider the square of length 2 with vertices at (1 , 1), (1 , 1), ( 1 , 1) and ( 1 , 1). Then the function is continuous everywhere outside the square, and everywhere inside the square. On the boundary, it is discontinuous everywhere except the points (0 , 1), (1 , 1) and (1 , 0). 3. Show that if f and g are continuous functions in an open subset U R n , then max( f,g ), min( f,g ), and  f  are also continuous; show that the converses do not hold, that is if max( f,g ) is continuous, it is not necessarily the case that f and g are continuous, and similarly for min( f,g ) and  f  . Solution. We will first show that  f  is continuous. This is clear from the fact that  f ( x )    f ( y )   f ( x ) f ( y )  . We obtain this because (  f ( x )    f ( y )  ) 2 =  f ( x )  2 +  f ( y )  2 2  f ( x )  f ( y )  f ( x ) 2 + f ( y ) 2 2 f ( x ) f ( y ) = { f ( x ) f ( y ) } 2 which immediately gives  f ( x )    f ( y )   f ( x ) f ( y )  . Then, by the definition of continuity of the function f , if we fix a point a , then we know, > 0, > 0 such that  x a  < =  f ( x ) f ( a )  < . But from what we just proved, it is easy to see that  f ( x )  f ( y )   f ( x ) f ( y )  < , so it follows that  f  is also continuous at a . Since f is continuous everywhere, it follows that  f  is also continuous everywhere. Now, we know that if f and g are continuous functions, then so are f + g and f g . From our proof above, it also follows that  f g  is continuous. But then, by adding the three continuous functions, we get f + g +  f + g  to be a continuous function. Scaling this down by a factor of 2, we get h 1 ( x ) = f ( x ) + g ( x ) +  f ( x ) g ( x )  2 3 is a continuous function. But this function is clearly nothing but max( f,g ). So it follows that max( f,g ) is a continuous function too. A very similar proof is immediate for min( f,g ) = h 2 ( x ), where h 2 is given by h 2 ( x ) = f ( x ) + g ( x )  f ( x ) g ( x )  2 We have thus just proved that if f and g are continuous, then so are max( f,g ), min( f,g ), and  f  ....
View
Full
Document
This note was uploaded on 05/03/2010 for the course MAT 218 at Princeton.
 '09
 RobertC.Gunning
 Sets

Click to edit the document details