Krantz S.G.'s A Guide to Functional Analysis PDF

By Krantz S.G.

ISBN-10: 0883853574

ISBN-13: 9780883853573

ISBN-10: 1614442134

ISBN-13: 9781614442134

This e-book is a short yet particular and cautious creation to the topic of practical research. It covers the fundamental subject matters that may be present in a uncomplicated graduate research textual content. however it additionally covers extra subtle issues corresponding to spectral idea, convexity, and fixed-point theorems. a distinct function of the e-book is that it features a nice many examples or even a few functions. It concludes with an announcement and facts of Lomonosov's dramatic outcome approximately invariant subspaces

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Extra info for A Guide to Functional Analysis

Example text

For the converse, suppose it is known that SN f ! T /. Then the uniform boundedness principle tells us that there is a constant C so that kSN kLp ;Lp Ä C . So we have a necessary and sufficient condition for Lp convergence of Fourier series. A classical calculation, amounting mainly to algebraic trickery (see [KRA3]), in fact reduces the question of the uniform bound on the operator norms of the SN to a single bound on a single operator called the Hilbert transform. t/ dt : f 7 ! x t The Hilbert transform is arguably the most important linear operator in all of analysis, and this discussion gives an indication of one of the reasons why.

R EMARK . It is actually quite difficult to give a “constructive” proof of the last result. And certainly functional analysis gives it to us rather easily. E XAMPLE . We shall take a moment to discuss the so-called closed graph theorem. This is an extremely useful criterion for telling when a linear operator is continuous. The statement is as follows: Theorem: Suppose that X and Y are Banach spaces. Let ƒ W X ! Y be a linear mapping. x; ƒx/ W x 2 Xg is a closed set in X Y . Then ƒ is continuous.

Define P x D x Qx. Then (3) is automatically true. Since Qx 2 x C E, it follows that P x 2 E. So P maps H into E. Next we show that hQx; ei D 0 for all e 2 E. We may assume that kek D 1, and we put z D Qx. The minimal property of Qx now shows that hz; zi D kzk2 Ä kz ˛ek2 D hz ˛e; z ˛ei for any scalar ˛. This simplifies to 0Ä ˛he; zi ˛hz; ei C j˛j2 : 40 3. Hilbert Space Take ˛ D hz; ei. This gives 0Ä jhz; eij2 ; so that hz; ei D 0. Hence Q maps H into E ?. Write x D xE C xE ? , with xE 2 E and xE ?

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A Guide to Functional Analysis by Krantz S.G.

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