Analysis Qual: Key Ideas and Problems To-Do
Will Feldman, Brent Nelson, Nick Cook, Alan Mackey
(Now with more Folland Problems!)
(Red is To-Do)
Spring 1970:
R2. f Lebesgue measurable and continuous at 1. g is in L^1. Evaluate the limit as n goes to
infinity of the integral from -n to n of f(1+x/n^2)g(x)dx.
Pull the indicator function 1_[-n,n] inside and note that for large enough n the integrand is
dominated by 2f(1)g(x).
R3. Find the extreme points of the unit ball in L^2([0,1]).
That is, find points h in the closed ball, such that h=.5(f+g) for f, g in the closed ball implies
h=f=g.
Write out the norm of h=.5(f+g) and Cauchy-Schwarz, then use ||f||,||g||<=1 to bound ||h||<=1.
Then look back at conditions for equality in those inequalities, and they imply f=g=h.
Fall 2001:
R1. (a) 1,-1,0,0,0.
... new line 0, 1, -1, 0.
.. etc. (b) Tonelli
R2. Closed and bounded. b) determinant continuous and maps to {+1,-1}
R4. 245B TA Section (density argument)
R5. Consider number of faces, vertices and edges of unit ‘ball’ polytopes
R6. 245b homework (linear functional mapping sequences to their limit)
C3. (b) induction
C4. See Sp02.11
C5. (a) uniform continuity
(b) Sequence of coefficients is square summable: use orthogonality to write L2 norm of
F(theta,r) for r fixed as sum |a_n|^2r^2n use mct as r goes up to 1 and continuity of F on the
closed disk.
Winter 2002:
1. Consider a sequence of “triangle” functions: f_n is zero on [1/n,1] and from [0,1/n] is a triangle
of height 1 with peak at 1/2n. Then f_n converges to zero everywhere but not uniformly.
[Counterexampled Prop 1.15), show induced measure is abs continuous w.r.t to Lebesgue
measure and a] (Not sure this is the right number)
3. Apply Fubini to LHS. Must verify integrand is integrable using Tonelli.
4. Define a premeasure by mu( (a,b])=F(b)-F(a) (Follanpply Radon-Nikodym
5. 245b discussion: pf by contradiction, consider preimages of diadic annuli, infinitely many
of which must have positive measure. Bound ||fg||_2 below by divergent sum for a carefully
chosen g. (can also use operator stuff, Closed Graph theorem)
6. Use a desnity argument
7. First consider X=[0,\infty). Since X has both sets of arbitrarily large and arbitarily small
measure, given any p*, L^p* is not contained in L^q for any q>p*, and is not contained in any L^r
for any r<p*.
So that means we have f_n in L^p* such that f_n is not in L^{p*+1/n} So we can make an f as
1