 Solution

ID: past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2007/a/1/d/solution

c_{k} = (2 π)^{- 1} \int_{- π}^{π} u (t) e^{- ik t} d t

- n, \dots, n

is therefore a convolution with

D_{n} (θ) = \sum_{k = - n}^{n} e^{ik θ} = \frac{s i n (( n + 1/2 ) θ )}{s i n ( θ /2 )} .

The last expression follows by summing a finite geometric series, with its removable value

2 n + 1

at zero. Periodicity and the evenness of

D_{n}

give

s_{n} (x) = \frac{1}{2 π} \int_{- π}^{π} u (x + θ) D_{n} (θ) d θ .

This also proves linearity and that the image lies in

F_{n}

. We use the continuous periodic function space, consistent with the premise and the displayed maximum norm.

The convolution formula gives the upper bound

∥ S_{n} u ∥_{\infty} \leq \frac{∥ u ∥ _{\infty}}{2 π} \int_{- π}^{π} ∣ D_{n} (θ) ∣ d θ .

To prove sharpness within continuous functions, take

u_{ϵ} (θ) = \frac{D _{n} ( θ )}{D _{n} ( θ ) ^{2} + ϵ ^{2}} .

It is continuous and periodic, with norm at most one. At

x = 0

, dominated convergence gives

(S_{n} u_{ϵ}) (0) \to (2 π)^{- 1} \int ∣ D_{n} ∣

. Hence that upper bound is the actual operator norm. Evenness yields

∥ S_{n} ∥ = \frac{1}{π} \int_{0}^{π} \frac{sin (( n + 1/2 ) θ )}{sin ( θ /2 )} d θ .

The approximating continuous signs avoid assuming that the discontinuous sign of the kernel belongs to the domain space.

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