Kenneth Kuttler

17.7. EXERCISES 421

both exist at every point. The above conditions rule out functions where the slopetaken from either side becomes infinite. Justify the following assertions and eventu-ally conclude that under these very reasonable conditions

limn→∞

Sn f (x) = ( f (x+)+ f (x−))/2

the mid point of the jump. In words, the Fourier series converges to the midpoint ofthe jump of the function.

Sn f (x) =∫

−π

f (x− y)Dn (y)dy∣∣∣∣Sn f (x)− f (x+)+ f (x−)2

∣∣∣∣=

∣∣∣∣∫ π

−π

(f (x− y)− f (x+)+ f (x−)

)Dn (y)dy

∣∣∣∣=

∣∣∣∣∣∫

0 f (x− y)Dn (y)dy+∫

0 f (x+ y)Dn (y)dy−∫

0 ( f (x+)+ f (x−))Dn (y)dy

∣∣∣∣∣≤∣∣∣∣∫ π

0( f (x− y)− f (x−))Dn (y)dy

∣∣∣∣+ ∣∣∣∣∫ π

0( f (x+ y)− f (x+))Dn (y)dy

∣∣∣∣Now apply some trig. identities and use the result of Problem 18 to conclude thatboth of these terms must converge to 0.

20. ↑Using the Fourier series obtained in Problem 11 and the result of Problem 19 above,find an interesting formula by examining where the Fourier series converges whenx = π/2. Of course you can get many other interesting formulas in the same way.Hint: You should get

Sn f (x) =n

∑k=1

2(−1)k+1

ksin(kx)

21. Let V be an inner product space and let K be a convex subset of V . This means thatif x,z∈K, then the line segment x+ t (z−x) = (1− t)x+ tz is contained in K for allt ∈ [0,1] . Note that every subspace is a convex set. Let y ∈ V and let x ∈ K. Showthat x is the closest point to y out of all points in K if and only if for all w ∈ K,

Re⟨y−x,w−x⟩ ≤ 0.

In Rn, a picture of the above situation where x is the closest point to y is as follows.

Kw θ

yx•

The condition of the above variational inequality is that the angle θ shown in thepicture is larger than 90 degrees. Recall the geometric description of the dot productpresented earlier. See Page 33.

17.7. EXERCISES 42120.21.both exist at every point. The above conditions rule out functions where the slopetaken from either side becomes infinite. Justify the following assertions and eventu-ally conclude that under these very reasonable conditionslim Syf (s) = (f+) + £6) /2the mid point of the jump. In words, the Fourier series converges to the midpoint ofthe jump of the function.Sif (x) = f° F@-y)Pulo)aysup (a) - FD EEO)~ [ (ro-y- Here) D,(v)ahJo f (@—y)Dn (y) dy + So f (x+y) Dn (y) dy— Jo (f (x+) + f (x—)) Dn (y) dySs i" (F(8-9) = F06-))Du(o) | + [ (f (x+y) —f (8+)) Dn (y) dyNow apply some trig. identities and use the result of Problem 18 to conclude thatboth of these terms must converge to 0.+Using the Fourier series obtained in Problem 11 and the result of Problem 19 above,find an interesting formula by examining where the Fourier series converges whenx = 7/2. Of course you can get many other interesting formulas in the same way.Hint: You should get2 (-1)"!Sif (x) = y a sin (kx)k=1Let V be an inner product space and let K be a convex subset of V. This means thatif x,z € K, then the line segment x +1 (z—x) = (1 —t)x+1z is contained in K for allt € [0,1]. Note that every subspace is a convex set. Let y € V and let x € K. Showthat x is the closest point to y out of all points in K if and only if for all w € K,Re (y —x,w—x) <0.In R", a picture of the above situation where x is the closest point to y is as follows.The condition of the above variational inequality is that the angle 9 shown in thepicture is larger than 90 degrees. Recall the geometric description of the dot productpresented earlier. See Page 33.