 Solution

ID: past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2010/a/2/d/solution

For

n \geq 1

ϕ_{n}

P_{n - 1}

, so

(u, ϕ_{n}) = (u - p_{n - 1}^{*}, ϕ_{n})

∣ (u, ϕ_{n}) ∣ \leq E_{n - 1} \int_{0}^{1} x^{2} ∣ ϕ_{n} ∣ d x \leq E_{n - 1} (\int_{0}^{1} x^{2} d x)^{1/2} ∥ ϕ_{n} ∥_{2} = E_{n - 1} / 3 .

Using the tail identity from part (b) and shifting the index,

∥ u - p_{n} ∥_{2}^{2} \leq \frac{1}{3} r = n \sum \infty E_{r}^{2} .

For the stated power law and

n > 1

, the decreasing-function integral test gives

\sum_{r = n}^{\infty} r^{- 2 a} \leq \int_{n - 1}^{\infty} x^{- 2 a} d x = \frac{( n - 1 ) ^{1 - 2 a}}{2 a - 1} .

Consequently

∥ u - p_{n} ∥_{2} \leq \frac{K}{3} \frac{( n - 1 ) ^{- a + 1/2}}{2 a - 1} .

The assumption

a > 1

is sufficient; this summation estimate itself only needs

a > 1/2

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