Part CCSIR NET June 2025T-has-norm-exactly-1-so-it-is-not-a-contraction-but-its-square-is

T has norm exactly 1 so it is not a contraction but its square is

Let C[0, 1] be the $R -$ vector space of real valued continuous functions equipped with the norm ‖f‖ = $s u p_{x \in [0, 1]}$ |f(x)|. Let T : C[0, 1] → C[0, 1] be defined as $T (f) (x) = \int_{0}$ ˣ f(t)dt, for x ∈ [0, 1]. Let $T^{n} = T$ ∘ T ∘ ⋯ ∘ T (n times). Which of the following statements are true?

A.There exists $α \in (0, 1)$ such that for all f, g ∈ C[0, 1], ‖T(f) − T(g)‖ $\leq α$ ‖f − g‖.
B.There exists $α \in (0, 1)$ such that for all f, g ∈ C[0, 1], ‖ $T^{2} (f) - T^{2} (g)$ ‖ $\leq α$ ‖f − g‖.✓
C.The set {f ∈ C[0, 1] : T(f) = f} is a singleton set.✓
D.‖ $T^{n}$ ‖ $\to \infty$ as $n \to \infty$ .

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

Related counterexample: Completeness is a topological property

The chapter behind this: Completeness, Banach fixed point and Baire category — free to read

From Metric Spaces › Completeness and Baire category

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