 Solution

ID: past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2008/a/3/b/solution

K

has one sign, the weighted mean-value theorem applied to the continuous derivative

u^{(n + 1)}

gives

\int_{0}^{1} K (t) u^{(n + 1)} (t) d t = u^{(n + 1)} (η) \int_{0}^{1} K (t) d t

for some

η \in [0, 1]

. One may prove this by bounding the weighted integral between the minimum and maximum derivative values times the total weight, then using the intermediate value theorem. If the total weight is zero, the fixed-sign kernel is zero almost everywhere and the conclusion is immediate.

u (x) = x^{n + 1}

to get

X [x^{n + 1}] = (n + 1)! \int_{0}^{1} K

. Consequently

X [u] = \frac{u ^{(n + 1)} ( η )}{( n + 1 )!} X [x^{n + 1}] .

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