This exam is closed book, no books, papers or recording devices permitted. You may use theorems from class, or the book, provided you can recall them correctly.
With short solutions and comments. I hope I did not damage too many delicate egos with this final!
Suppose and for all Show that for all with implies that almost everywhere with respect to Lebesgue measure.
Generally well done.
Ans:- From a standard result, a.e. in for every The subset of on which is therefore the union of a countable collection of sets of measure zero, so also of measure zero. Hence almost everywhere in
Suppose is a bounded and self-adjoint operator on a Hilbert space show that if for all then
Only one person got this right. I did not say compact, so you cannot use the spectral theory of compact operators. No, all bounded operators are not compact etc.
Ans:- For any expanding using sesquilinearity (and linearity of gives
Give an example of an element of which is not the Fourier transform of an element of
Of course, I confused people by giving a harder version of this on the practice final.
Ans. If the Fourier transform of is in then it must be the inverse Fourier transform of an function, hence continuous. Thus it suffices to find an function which is not continuous, say the characteristic function of
If show that
Generally well done although quite a few people tried to use Fourier series.
Ans:- By the translation-invariance of Lebesgue measure and countable additivity, the terms, for in the series satisfy
Show that there is no function satisfying
Mostly people got this one.
Ans:- If there is such a function its real part also satisfies the condition, so we can assume it is real. Then the integral in () is times the imaginary part of the th Fourier coefficient, thus which series diverges, violating Bessel's inequality.
Show that the function is in and that its Fourier transform is infinitely differentiable. Is a Schwartz function?
Many people thought this function was Schwartz, it is not infinitely differentiable and that is one of the requirements!
Ans: Certainly and we know (if you like from class) that for every Since follows that is continuous for every No, it is not Schwartz.
Suppose is such that and the Fourier transform is such that Show that there exists a function such that
Many people got this one at least more-or-less right. However, many of the arguments to justify the integration by parts were dubious to say the least.
Ans:- Since is given, we only need to find satisfying
If show that
One can argue directly as I did in the practice exam and show that has equismall tail with respect to the Fourier basis when Or
Ans:- If one sets when and zero otherwise one gets a function in sucht that
Show that there exists an element which has but for every
Rather a lot of wild things said here.
Ans:- If then its Fourier transform is Schwartz and conversely. We also know that
Richard B. Melrose 2004-05-24