Math 5b Homework Solutions January 5, 5 Problem (R-S, II.) () For the R case, we just expand the right hand side and use the symmetry of the inner product: ( x y x y ) = = ((x, x) (y, y) (x, y) (y, x) (x, x) (y, y) (x, y) (y, x)) = = ((x, y) (y, x)) = (x, y). For the other case, we again expand the right side, using the relation we just proved: ( x y x y ) i ( x iy x iy ) = = ((x, y) (y, x)) i ((x, iy) (iy, x)) = = (x, y) (y, x) i (x, y) i (y, x) = (x, y). () If the norm comes from an inner product, then we have x y x y = = (x, x) (y, y) (x, y) (y, x) (x, y) (y, x) = ( x y ). Now suppose that the norm satisfies the parallelogram law. Assume the field is C (the argument is similar for R), and define the inner product via the polarization identity from part (). If x, y, z V, we have that (writing x y = x yz yz and similar
for x z): (x, y) (x, z) = ( x y x z x y x z ) i ( x iy x iz x iy x iz ) = = ( x y z y z x y z y z ) i ( x iy z iy z x iy z iy z ) = = ( x y z y z y z x y z ) i ( x iy z iy z iy z x iy z ) = = ( x y z x x (y z) x ) i ( x i(y z) x x i(y z) x ) = (x, y z). This holds for all x, y, z, so, in particular, (x, ny) = n(x, y), and so if r = p/q Q, then (x, ry) = r(x, y). Moreover, by the polarization identity, (x, iy) = ( x iy x iy ) i ( x y x y ) = i(x, y). Combining these two results gives that if α Q iq, then (x, αy) = α(x, y). If α C, then there are α n Q iq, α n α, so then α n y αy, so (x, α n y) (x, αy) because all the norms above must converge. This gives us that α(x, y) = lim α n (x, y) = lim(x, α n y) = (x, αy), so (x, y) is linear. Observe that because i(x iy) = x iy, we have that (y, x) = (x, y), and that (x, x) = ( x ) i ( i x i x ) = x, so this shows that the norm is induced by (, ) and that it is also positive definite, so it is an inner product.
(3) Let f = χ [,/], g = χ [/,] L p (R), so then f g = f g =, and f = g = ( )/p, so the parallelogram law does not hold unless p =. (For p =, the same examples work, but the norms are slightly different.) Let a = (/, /,,,...), b = (/, /,,,...) l p, then a = b = ( ( )p ( )p) /p, while a b = a b =, so the parallelogram law does not hold unless p =. Problem (R-S, II.8) Fix x, so (i) and (iii) show B(x, ) is a continuous linear functional on H, so there is some x H such that B(x, y) = (x, y) for all y. Let Ax = x. By construction, for all x, y, we have B(x, y) = (Ax, y). We want to show that A is a continuous linear transformation with the right norm and that it is unique. Let x, y, z H, α C. Then (A(x y), z) = B(x y, z) = B(x, z) B(y, z) = (Ax, z) (Ay, z) = (Ax Ay, z), holds for all z, so A(x y) = Ax Ay. Similarly, (A(αx), y) = B(αx, y) = ᾱb(x, y) = ᾱ(ax, y) = (αax, y), so A(αx) = αax, so A is linear. For a fixed x, we choose y = Ax, then B(x, Ax) = (Ax, Ax) = Ax, but B(x, Ax) C x Ax, so Ax C x, so A is a continuous linear transformation. We just showed that A C for all C such that B(x, y) C x y, but we also have that B(x, y) = (Ax, y) Ax y A x y, so A has the right norm. Finally, suppose that T : H H is another such bounded linear transformation, then B(x, y) = (Ax, y) = (T x, y) for all x, y. Then A T is a bounded linear transformation, and ((A T )x, y) = for all x, y. Choose y = (A T )x, then (A T )x =, so Ax = T x, so A is unique. Problem 3 Suppose that f C (S ) with f(y) =. We claim that (T f)(y) =. Indeed, note that f (x) = f(x) y x = f (ty ( t)x)dt is smooth, since we can differentiate with respect to x under the integral. Then f(x) = (e ix e iy ) g(x) where g(x) = xy f e ix e iy (x) is smooth. Thus T f(y) = (e iy e iy )T g(y) =. We claim now that T is a multiplication operator. Indeed, consider the constant function g(x) = f(y), then f(x) g(x) vanishes at y, so T (f g)(y) =, and thus, for each y, T f(y) = f(y)t ()(y), 3
i.e., T f(x) = φ(x)f(x) for all x, where φ(x) = (T )(x). Note that because the constant function is a smooth function and T takes values in C (S ), we must have that φ C (S ) as well. We finally claim that φ is constant. To do this we use that T commutes with differentiation. Because φ(x) = (T )(x), we have φ (x) = d (T )(x) = T dx so that φ must be constant. This finishes the proof. Problem. Integration by parts shows that (e n, f ) = π π = π π = π π This holds for all n, proving the claim.. Let f C (S ). Then e inx f (x)dx ( ) d dx einx (e n, f) = (e, f) n Z ( ) d dx (x) =, f(x)dx f(x)e inx π ine inx f(x)dx = in(e n, f). (e, f) n Z, n ( n n (e n, f ) ) f n L C f C C f L C f C (S ). which proves Ff l (Z) and Ff l (Z) C f C (S ) so that F is continuous. 3. We start by noting that e n C (S ) = π for all n. This implies that for {a n } l (Z), the series a n e n converges absolutely (and so uniformly), as a n e n (x) {a n } l (Z). π Note that this inequality also shows that the map is continuous. Together with the previous part, we may conclude that the Fourier series of f C (S ) converges uniformly.
. Note that the previous part shows that both f = F {c n } and g = F {inc n } are continuous functions, defined by uniformly convergent series. Note further that for each partial sum f N = N N c ne n, we have that f N = N N inc ne n. We know already that the sequence f N converges uniformly to F {inc n }. We claim now that this is enough to show that f = g and so that f C (S ). Indeed, we observe that for x [, π], f(x) = lim N f N(x) = lim N x f N(y)dy. f N converges uniformly to g and S is compact, so we may conclude by the dominated convergence theorem that f(x) = x g(y)dy. In particular, the fundamental theorem of calculus then tells us that f (x) = g(x) and so f C (S ). This also shows that d F {a dx n } = F {ina n }. 5. We start by calculating, for f L (S ), If {c n } l (Z), then F ( e ix f ) n = (e n, e ix f) = π π = π π F {c n } = π cn e inx = e inx e ix f(x)dx e i(n)x f(x)dx = (e n, f) = (Ff) n. eix π cn e i(n)x = e ix π cn e inx = e ix F {c n }. 6. Part. of this problem shows that if f is in C, then Ff lies in l. In particular if f C (S ), then f (k) C (S ) for any k and so {(in) k (Ff) n } l (Z) for any k. In particular, this is bounded and so Ff s(z). Similarly, part. of this problem shows that if {nc n } l (Z) then F {c n } C (S ). If {c n } s(z), then {n(in) k c n } l (Z) for any k. Thus F {(in) k c n } C (S ) for any k. By iterating problem part., we see that F {(in) k } = dk F {c dx k n }, and so F {c n } C k (S ) for any k, i.e., in C (S ). 7. By parts.,., 5., and 6. of this problem, we know that F F : C (S ) C (S ) commutes with multiplication by e ix and with differentiation. By problem., we conclude that F F = cid, where c is a constant. To determine the value of c, we consider the function. c n = (F) n = (e n, ) = π π e inx = { π n = n. 5
We then observe that F F = π e =, so that c =. 8. F F = Id on C (S ) and is continuous on L. We know that C (S ) is dense in L (S ), so it must equal the identify on L. 6