GATE 2020 MA – Question 23
Let f∈C[0,1] satisfy $|f(x)-f(y)|\leq M|x-y|$ for all x,y and some M > 0. Set $f_n(x)=f(x^{1+1/n})$ and $S=\{f_n:n\in\mathbb N\}$ in the supremum norm. Then
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Correct answer: (A) The closure of S is compact.
Explanation
The maps $x^{1+1/n}$ converge uniformly to x on [0,1]. The Lipschitz bound gives $\|f_n-f\|_\infty\leq M\sup_x|x^{1+1/n}-x|\to0$. A convergent sequence together with its limit is compact, so the closure of S is compact. S need not contain its limit, for example f(x) = x, so S itself need not be closed or compact. Thus A.