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Answer whether the following statements are True or False.

Let $\alpha$ be a positive real number, and let $f:(0,1) \rightarrow \mathbb{R}$ be a function such that $|f(x)-f(y)| \leq$ $|x-y|^{\alpha}$ for all $x, y \in(0,1)$. Then $f$ can be extended to a continuous function $[0,1] \rightarrow \mathbb{R}$.

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|f(x)-f(y)| ≤ |x-y|ᵃ is called a-Hölder condition when a>0.

This condition implies that f(x) is Hölder continuous. Every Hölder continuous function is Uniformly Continuous.

∴ f:(a,b)→ℝ is Uniformly Continuous function.

If {xₙ} is any Cauchy sequence in (a,b), then {f(xₙ)} is a Cauchy sequence as f is uniformly continuous.

Consider xₙ = a+1/n, xₙ is Cauchy, hence f(xₙ) is also Cauchy and converges to some α, since ℝ is complete, and

Consider yₙ = b-1/n, yₙ is Cauchy, hence f(yₙ) is also Cauchy and converges to some β, since ℝ is complete.

Define g:[a,b]→ℝ as g(a) = α, g(b) = β, g(x) = f(x) when a<x<b.

It is not hard to prove the continuity of g at a and b using uniform continuity of f. So, the statement is ‘’true’’.
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