 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-61/1/2/b/solution

The indexing of the Fejér sums gives

(n + m) σ_{n + m} (f) = \sum_{j = 0}^{n + m - 1} s_{j} (f), n σ_{n} (f) = \sum_{j = 0}^{n - 1} s_{j} (f) .

Subtracting removes precisely the initial

n

v_{n, m} (f) = \frac{( n + m ) σ _{n + m} ( f ) - n σ _{n} ( f )}{m} .

For

n = 0

, omit the second term, so that no undefined

σ_{0}

∥ v_{n, m} (f) ∥_{\infty} \leq \frac{n + m}{m} ∥ σ_{n + m} (f) ∥_{\infty} + \frac{n}{m} ∥ σ_{n} (f) ∥_{\infty} \leq (1 + \frac{2 n}{m}) ∥ f ∥_{\infty} .

Hence the operator norm of the de la Vallée Poussin sum is at most $1 + 2 n / m$ .

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