 Solution

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

Let

(x_{n})

y_{n} = T x_{n}

. Since

∥ y_{n} - y_{m} ∥_{Y} \leq ∥ x_{n} - x_{m} ∥_{T},

(y_{n})

Y

. Write

y_{n} \to y \in Y

Use the bounded right inverse to remove the part seen by

T

w_{n} = x_{n} - S y_{n}, T w_{n} = T x_{n} - TS y_{n} = y_{n} - y_{n} = 0.

Thus

w_{n} \in W

, and the operator norm bound gives

∥ w_{n} - w_{m} ∥_{X} \leq ∥ x_{n} - x_{m} ∥_{X} + ∥ S ∥∥ y_{n} - y_{m} ∥_{Y} \leq (1 + ∥ S ∥) ∥ x_{n} - x_{m} ∥_{T} .

The assumed completeness of

W

supplies

w \in W

with

w_{n} \to w

∥ \cdot ∥_{X}

. Put

x = w + S y \in X

. The linearity of

T

and the right inverse identity give

T x = y

. Therefore

∥ x_{n} - x ∥_{T} \leq ∥ w_{n} - w ∥_{X} + (1 + ∥ S ∥) ∥ y_{n} - y ∥_{Y} ⟶ 0.

(X, ∥ \cdot ∥_{T}) is a Banach space .

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