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
−π
f (x− y)Dn (y)dy∣∣∣∣Sn f (x)− f (x+)+ f (x−)2
∣∣∣∣=
∣∣∣∣∫ π
−π
(f (x− y)− f (x+)+ f (x−)
2
)Dn (y)dy
∣∣∣∣=
∣∣∣∣∣∫
π
0 f (x− y)Dn (y)dy+∫
π
0 f (x+ y)Dn (y)dy−∫
π
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.