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Spring 2013 Analysis Prelim Solutions

This document contains summaries of 6 problems from a Spring 2013 analysis prelim exam. 1) The first problem shows that the orthogonal projection of a function g(x) onto a subspace V is given by subtracting a function w(x) in the orthogonal complement V^⊥ from g(x). It computes this projection for a specific example. 2) The second problem shows that if λ is not in the spectrum of an operator A, then λ is also not in the spectrum of approximating operators A_n as n goes to infinity. 3) The third problem computes a limit of an integral involving exponential and trigonometric functions, showing the limit is 0. 4

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Patrick Tam
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72 views3 pages

Spring 2013 Analysis Prelim Solutions

This document contains summaries of 6 problems from a Spring 2013 analysis prelim exam. 1) The first problem shows that the orthogonal projection of a function g(x) onto a subspace V is given by subtracting a function w(x) in the orthogonal complement V^⊥ from g(x). It computes this projection for a specific example. 2) The second problem shows that if λ is not in the spectrum of an operator A, then λ is also not in the spectrum of approximating operators A_n as n goes to infinity. 3) The third problem computes a limit of an integral involving exponential and trigonometric functions, showing the limit is 0. 4

Uploaded by

Patrick Tam
Copyright
© Attribution Non-Commercial (BY-NC)
We take content rights seriously. If you suspect this is your content, claim it here.
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Analysis Prelim Practice

Patrick Tam September 8, 2013

Spring 2013 Analysis Prelim Problem 1) Let U be the subspace spanned by the function h(x) = x. Then U is precisely V . We know that (U ) = U . Since U is nite dimensional, it is a closed subspace so
U = U = (U ) = V .

The orthogonal projection of g (x) onto V is given by subtracting a function w in V from g such that
f (x) = g (x) w(x) V.

We know that w(x) = ax for some a C. So we can write


f (x) = 1 ax.

To pick a such that f (x) V , we just provide the following computation.

0 1

xf (x)dx = 0
0 0 1

x(1 ax)dx = 0
1

x ax2 dx = 0 ax3 x2 2 3 |1 0 =0

1 a (0 0) = 0 2 3 3 a= . 2 3 Thus, f (x) = 1 2 x is in V . It is the closest element in V to g (x) = 1 since it is the orthogonal projection of g onto V .

Spring 2013 Analysis Prelim Problem 2) Our strategy will be to show that if / (A), then / 0 . Suppose / (A). Then I A is invertible and (I A)1 is a bounded linear operator by the inverse mapping theorem. Let M > 0 be a bound for (I A)1 . So we can write,
(I A)1 (I An ) I = (I A)1 (I An ) (I A)1 (I A) = (I A)1 ((I An ) (I A)) = (I A)1 (A An ) .

So

(I A)1 (I An ) = I + (I A)1 (A An ) = I (I A)1 (An A) .

Since An A in norm, there exists N N such that


||An A|| < 1 2M

for all n N . Let x B . Then


||(I A)1 (AN A) (x)||B M || (AN A) (x)||B M 1 1 ||x|| ||x||. 2M 2

1 . Using the power series expansion, we get Thus, ||(I A)1 (AN A) || 2

(I A)1 (I AN )

= I (I A)1 (AN A)

=
i=0

(I A)1 (AN A) .

Since ||(I A)1 (AN A) || Thus,


i=0

1 2

< 1, this series converges and is an inverse for (I A)1 (I AN ).


i

(I A)1 (AN A)
i=0 i

(I A)1 (I AN ) = I

and (I A)1 (AN A) (I A)1 is an inverse for I AN . So I AN is invertible with a bounded inverse by the inverse mapping theorem. Thus, / (AN ) = 0 and we conclude that 0 (A). Spring 2013 Analysis Prelim Problem 4) We want to show that
lim
0 0 2

+x

sin

1 x

dx = 0.

Lets denote the integrand by h(x) and let


f (x) =
2

+x 1 x .

and
g (x) = sin

Then f g = h. f L2 (0, ) since

0 2

|f (x)| dx =
0 2

du =
2 2

+x
2

dx = 1 1 2 dx =
2

1 du u2

| 2

1 u

1
2

= 1.

Also, g L2 (0, ) and we can compute

|g (x)|2 dx =
0

sin

1 x

sin2
0

1 x

dx =

. 2

So after using the linearity of integration and Holder inequality, we get:


||h||1 = || h||1 ||f ||2 ||g ||2 = 1
2

. 2

Thus,
||h||1

. 2

From here, it is clear that taking the limit as tends to 0 will give us 0 for our integral. Spring 2013 Analysis Prelim Problem 5) We just compute:
k | = | (in) f | = |(in)k | |f | = |n|k |f |. |fn n n n k

k }, is a sequence that Since f k is continous, it is in L2 (T) and its image under the Fourier transform, {fn tends to 0 as n tends to . Let > 0 and let us consider

|n|k .
k | tends to 0 as |n| tends to , we can pick N N such that Since |fn

|n|k |fn | <

for all n N . So |fn | < |n|k for all n N . This proves the rst half of the problem. For the second half of the problem, we consider a sequence {cn } such that |cn | is in O(|n|k1 ) for some > 0. Then there exists a function h on T such that hn = cn . Then |cn (in)k | is in O(|n|1 ). In particular, you can write {cn (in)k } as a product of two sequences, {an } and {bn }, such that cn (in)k = an bn for all n N and the absolute values of each of those two sequences is in O(|n|(1 )/2 ). It is clear that {|an |2 } and {|bn |2 } are both in O(|n|(1 ) ). So {an } and {bn } are the fourier transforms of two functions a and b in L2 (T). Since the inverse fourier transform converts multiplication to convolution, the convolution a b is then the kth derivative of h. Since a and b are both in L2 (T), a b C (T). So hk = a b is continuous. This means hk is continuous and thus h has k continuous derivatives and so h C k (T). Spring 2013 Analysis Prelim Problem 6)

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