Problem Set 3 Solution - Solutions to Problem set 3 15.3 By integration by parts we have 1 and 1 f(x sin nxdx = f(x cos nxdx = 1 n cos nxdf(x 1 n sin

# Problem Set 3 Solution - Solutions to Problem set 3 15.3 By...

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Solutions to Problem set 3 15.3. By integration by parts we have 1 π Z π - π f ( x ) sin nxdx = 1 Z π - π cos nxdf ( x ) , and 1 π Z π - π f ( x ) cos nxdx = - 1 Z π - π sin nxdf ( x ) , where the boundary terms cancel because f ( π ) = f ( - π ). The right hand sides give a bound of 1 V π - π f . 15.5 a) Since Riemann integrability only applies to bounded functions, it must be that f is square integrable. Pick P = { x 0 , x 1 , . . . , x n } a partition of [ - π, π ] so that U ( f, P ) - L ( f, p ) < 2 8 || f || . Then we can define a step function k with h = M i for each x between x i - 1 and x i . Then Z ( f - k ) 2 2 || f || | infty Z f - k = 2 4 . Thus we have || f - k || 2 2 . We can modify k to a continuous g with || f - g || 2 by changing h only on small intervals around each x j and setting g to be linear on those intervals. Intervals of length smaller than 2 8 n || f || suffice. b) Using part a), we may find a function g C ([ - π, π ]) with || f - g || 2 2 . Now we need only find a 2 π periodic continuous function h with || h - g || 2 2 . We can do this by finding a continuous function on R s η with || s ||