Download Elements of Functional Analysis by I. J. Maddox PDF

By I. J. Maddox

The second one version of this winning textbook, first released in 1970, keeps the goals of the 1st, particularly to supply a very introductory path in useful research, however the chance has been taken so as to add extra element and labored examples. the most alterations are whole revisons of the paintings on convex units, metric and topological linear areas, reflexivity and vulnerable convergence. extra fabric at the Weiner algebra of totally convergent Fourier sequence and on vulnerable topologies is integrated. a last bankruptcy contains hassle-free purposes of practical research to differential and essential equations.

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Let g be differentiable on an interval containing /(c+ h) for h sufficiently small, and suppose g’ is continuous at /(c). Then g o / is differentiable at c and Lagrange’s Mean Value Theorem and Related Functional Equations 36 Since g is differentiable, by Lagrange’s mean value theorem, we get g(f(h + c)) - g(f(c)) = g'{6) [/(c + h) - /(c)] for some Θ strictly between f(c + h) and /(c). Now / is differentiable at c, so lim /i- 0 f(c+ h ) —f(c) h =m As / is continuous, lim /(c + h) = f(c) and thus lim0 = /(c) (since Θ is strictly between f(c + h) and /(c)).

This completes the proof of the theorem. Rem ark 1 . 2 simplifies to a great extent. 25) for s = —t is symmetric in x and y. Thus using this symmetry one can conclude that h is an even function. 49) is additive. Rem ark 2 . 1, h(y) is undefined a ty = 0. Rem ark 3. We have seen in Chapter 1 that the Cauchy functional equation A(x + y) = A(x) + A(y) has discontinuous solutions in addition to the continuous solutions of the form A(x) = ax, where a is an arbitrary real constant. 25) has discontinuous solutions.

This following proof is due to Tucker (1997) and Velleman (1998) (see Horn (1998)). Start with a nonempty interval [xi, xq\on which / is differentiable and define 29 Lagrange's Mean Value Theorem Then y divides the interval [rci, x2] into two subintervals of length h = Observe that < m < max{mi,m2}, where f(y) - f(x 1) _ _ .................. m i = ---- s7---- _,nH and / t e ) - f(y) h m 2 ----= --------- It follows by the intermediate value theorem, that the function /(* + h)~ f(x) g„,T\ ( x ) -----------takes the value m somewhere on [ai,6j] such that / ( f c ) - / ( « i) m III.

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