 Solution

ID: past-exam-of-the-functional-analysis-i-course-of-the-university-of-oxford/2023/1/d/ii/solution

f_{n} (x) = x^{2} + n^{- 2} .

The estimate

a^{2} + b^{2} \leq a + b

for

a, b \geq 0

gives

0 \leq f_{n} (x) - ∣ x ∣ \leq \frac{1}{n}, ∥ f_{n} - ∣ \cdot ∣ ∥_{\infty} = \frac{1}{n} .

Thus

f_{n} \to ∣ \cdot ∣

∥ f_{n} - f_{m} ∥_{\infty} \leq n^{- 1} + m^{- 1}

X

, but its unique possible supremum norm limit is

x \mapsto ∣ x ∣

, which is not differentiable at zero. That limit is outside

X

, so

(C^{1} ([- 1, 1]), ∥ \cdot ∥_{\infty}) is incomplete .

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