 Solution

ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2014/1/b/v/solution

True. For every finite partial sum,

\sum_{j = 0}^{N} ∣ x_{j} ∣^{2} \leq (\sum_{j = 0}^{N} ∣ x_{j} ∣)^{2} \leq ∥ x ∥_{1}^{2} .

The nondecreasing partial sums on the left are bounded, so the square series converges. Hence

ℓ^{1} \subseteq ℓ^{2}

and

∥ x ∥_{2} \leq ∥ x ∥_{1} .

Applying this norm inequality to

x - y

proves

∥ f (x) - f (y) ∥_{2} \leq ∥ x - y ∥_{1}

. Thus the inclusion is a Lipschitz continuous map, in particular continuous, and it is well-defined on every element of l1 sequence space.

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