# HW 01 - Mathematics 218 Analysis in Several Variables Fall...

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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, ∃ δ > 0 such that | x- a | < δ = ⇒ | 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 | ....
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HW 01 - Mathematics 218 Analysis in Several Variables Fall...

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