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!

- Problem 1
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

- Problem 2
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

Setting implies that for all that is as an operator.- Problem 3
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

- Problem 4
If show that

converges to an element ofGenerally 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

Thus, the series is Cauchy in and hence converges by the completeness of- Problem 5
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.

- Problem 6
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.

- Problem 7
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

and then will solve the actual problem. If we set as we can with uniquely determined, then so

where by definition is the function with Fourier transform equal to - in by assumption.- Problem 8
If show that

defines a compact operatorOne 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

Thus, from result in class this operators is Hilbert-Schmid, hence compact.- Problem 9
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

Thus we just have to arrange that and ALL its derivatives vanish at the origin. We do know that there is a non-trivial Schwartz function which vanishes outside for instance. Taking this as the Fourier transform of and then choosing the positive constant so that solves the problem.

Richard B. Melrose 2004-05-24