Solution

X

A \subseteq X

is closed, and

f : A \to [- M, M]

is continuous, then there is a continuous

F : X \to [- M, M]

with

F ∣_{A} = f

We first prove an approximation lemma. Given continuous

h : A \to [- r, r]

, the closed subsets

A_{-} = {x \in A : h (x) \leq - r /3}, A_{+} = {x \in A : h (x) \geq r /3}

are disjoint and closed in

X

. By the Urysohn lemma, there is a continuous

g : X \to [- r /3, r /3]

equal to

- r /3

A_{-}

and

r /3

A_{+}

. Then

∣ h - g ∣ \leq \frac{2 r}{3} on A .

Starting with

h_{0} = f

and

r_{0} = M

, apply the lemma recursively to obtain continuous

g_{j} : X \to R

such that

∥ g_{j} ∥_{\infty} \leq \frac{r _{j}}{3}, h_{j + 1} = h_{j} - g_{j} ∣_{A}, ∥ h_{j + 1} ∥_{\infty} \leq r_{j + 1},

where

r_{j} = M (2/3)^{j}

. The series

F = \sum_{j = 0}^{\infty} g_{j}

converges uniformly by the Weierstrass test, so its sum is continuous. On

A

, its remainder after

j

terms is

h_{j + 1}

, which tends uniformly to zero; hence

F ∣_{A} = f

. The bounds also keep

F

[- M, M]

after the standard endpoint-preserving version of the construction, proving the theorem.

Solved by gpt-5.6-sol high.

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