 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/22f/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Write

T_{v} = T (v)

. We prove that

T

has a closed graph. Suppose

v_{n} ⟶ v in V, T_{v_{n}} ⟶ f in V^{*} .

Put

u_{n} = v_{n} - v

and

g = f - T_{v}

. Then

u_{n} \to 0

and

T_{u_{n}} \to g

V^{*}

Fix

w \in V

and

t \in R

. Positivity and linearity give

0 \leq T_{u_{n} + tw} (u_{n} + tw) = T_{u_{n}} (u_{n}) + t (T_{u_{n}} (w) + T_{w} (u_{n})) + t^{2} T_{w} (w) .

Because

(T_{u_{n}})

converges in the continuous dual space, its norms are bounded, so

∣ T_{u_{n}} (u_{n}) ∣ \leq ∥ T_{u_{n}} ∥ ∥ u_{n} ∥ ⟶ 0.

Also

T_{u_{n}} (w) \to g (w)

, while

T_{w} (u_{n}) \to 0

because the fixed functional

T_{w}

is continuous. Passing to the limit yields

0 \leq t g (w) + t^{2} T_{w} (w)

for every real

t

. If

g (w) \neq = 0

, a sufficiently small

t

of the opposite sign makes the right-hand side negative. Hence

g (w) = 0

for every

w

, so

g = 0

and

f = T_{v}

The graph is therefore closed. The closed graph theorem now proves that

T : V \to V^{*}

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