Let \(M\) be the space of real-valued functions \(f\) on \([0,1]\) for which \(f\) and \(f'\) are continuous, with norm \[ \|f\|=\sup_{0\le x\le1}|f(x)|+\sup_{0\le x\le1}|f'(x)|. \] Which subsets of \(M\) are compact?
Solution
For a subset \(K\subseteq M\), write \[ K' \coloneqq \{f':f\in K\} \subseteq C[0,1]. \]
The compact subsets of \(M\) are exactly the subsets \(K\) satisfying all three conditions:
- \(K\) is closed in \(M\);
- \(K\) is bounded in the given norm;
- \(K'\) is equicontinuous on \([0,1]\).
1. If \(K\) is compact in \(M\), then \(K\) is closed and bounded in \(M\).
Every compact subset of a metric space is closed and bounded.
2. If \(K\) is compact in \(M\), then \(K'\) is equicontinuous.
The derivative map \[ D:M\longrightarrow C[0,1], \qquad D(f)=f', \] is continuous when \(C[0,1]\) has the uniform norm, because \[ \norm{f'-g'}_\infty \leq \norm{f-g}. \] Therefore \(K'=D(K)\) is compact in \(C[0,1]\). By the Arzelà–Ascoli theorem, every compact family of continuous functions on \([0,1]\) is equicontinuous.
3. Conversely, suppose \(K\) is closed and bounded in \(M\) and \(K'\) isequicontinuous. Then every sequence \((f_n)\) in \(K\) has a subsequence forwhich \((f_n')\) converges uniformly on \([0,1]\).
Boundedness of \(K\) in \(M\) gives a constant \(C\) such that \[ \norm{f_n'}_\infty\leq C \] for every \(n\). Thus the derivative family is uniformly bounded, and it is equicontinuous by hypothesis. The Arzelà–Ascoli theorem gives a uniformly convergent subsequence of \((f_n')\).
4. After passing to a further subsequence, there are\(a\in\RR\) and \(g\in C[0,1]\) such that\[ f_n(0)\longrightarrow a \]and\[ f_n'\longrightarrow g \]uniformly.
3. Boundedness of \(K\) also gives\[ \abs{f_n(0)} \leq \norm{f_n}_\infty \leq \norm{f_n} \leq C. \]Hence the real sequence \((f_n(0))\) has a convergent subsequence. Passingto it preserves the uniform convergence of the derivatives.
5. Define\[ f(x) \coloneqq a+\int_0^x g(t)\,dt. \]Then\[ f_n\longrightarrow f \]in the norm of \(M\).
The function \(f\) is continuously differentiable and \[ f'=g. \] For every \(x\in[0,1]\), \[ f_n(x) = f_n(0)+\int_0^x f_n'(t)\,dt. \] Therefore \[ \begin{aligned} \abs{f_n(x)-f(x)} &\leq \abs{f_n(0)-a} +\int_0^x\abs{f_n'(t)-g(t)}\,dt\\ &\leq \abs{f_n(0)-a} +\norm{f_n'-g}_\infty. \end{aligned} \] Taking the supremum over \(x\) shows \[ \norm{f_n-f}_\infty\longrightarrow0. \] Step 4 also gives \[ \norm{f_n'-f'}_\infty = \norm{f_n'-g}_\infty \longrightarrow0. \] Thus \[ \norm{f_n-f} = \norm{f_n-f}_\infty +\norm{f_n'-f'}_\infty \longrightarrow0. \]
6. The limit \(f\) from step 5 belongs to \(K\).
The subsequence consists of points of \(K\) and converges to \(f\) in \(M\). Since \(K\) is closed in \(M\), one has \(f\in K\).
7. Under the three stated conditions, \(K\) is compact.
Steps <1>3–<1>6 show that every sequence in \(K\) has a subsequence converging in \(M\) to a point of \(K\). Thus \(K\) is sequentially compact. Because \(M\) is a metric space, sequential compactness is equivalent to compactness.
8. Q.E.D.
Steps <1>1–<1>2 prove that compactness implies the three conditions, and step 7 proves the converse.