The Arzelà–Ascoli theorem asserts that a sequence \(\{f_n\}\) of continuous real-valued functions on a metric space \(\Omega\) is precompact (i.e., has a uniformly convergent subsequence) if
\(\Omega\) is compact,
\(\sup_n\|f_n\|<\infty\), where \(\|f_n\|=\sup\{|f_n(x)| : x\in\Omega\}\),
the sequence is equicontinuous.
Give examples of sequences which are not precompact such that: (i) and (ii) hold but (iii) fails; (i) and (iii) hold but (ii) fails; (ii) and (iii) hold but (i) fails. Take \(\Omega\) to be a subset of the real line. Sketch the graph of a typical member of the sequence in each case.
Solution
1. Conditions (i) and (ii) can hold while (iii) fails, without precompactness.
Take \[ \Omega=[0,1], \qquad f_n(x)=x^n. \] The space \(\Omega\) is compact, and \[ \norm{f_n}_\infty=1 \] for every \(n\), so (i) and (ii) hold.
The family is not equicontinuous at \(x=1\). Indeed, if equicontinuity held there, then for \(\varepsilon=1/2\) there would be \(\delta>0\) such that \[ \abs{x-1}<\delta \quad\Longrightarrow\quad \abs{x^n-1}<\frac12 \] for every \(n\). Choose any \[ x\in\bigl(\max\{0,1-\delta\},1\bigr). \] Then \(x\in[0,1)\) and \(\abs{x-1}<\delta\). Since \(x^n\to0\), the displayed inequality fails for all sufficiently large \(n\).
Moreover, every subsequence \((f_{n_k})\) has the same pointwise limit \[ f(x)= \begin{cases} 0,&0\leq x<1,\\ 1,&x=1. \end{cases} \] This limit is discontinuous. A uniformly convergent subsequence of continuous functions would have a continuous pointwise limit, so no such subsequence exists.
The graph of \(f_n\) runs from \((0,0)\) to \((1,1)\) and, as \(n\) grows, stays increasingly close to the \(x\)-axis until very near \(x=1\).
2. Conditions (i) and (iii) can hold while (ii) fails, without precompactness.
Again take \[ \Omega=[0,1], \] but now let \[ f_n(x)=n \] for every \(x\in[0,1]\). The domain is compact. The sequence is equicontinuous because \[ \abs{f_n(x)-f_n(y)}=0 \] for all \(x,y\) and all \(n\).
On the other hand, \[ \norm{f_n}_\infty=n, \] so (ii) fails. If a subsequence converged uniformly, then it would be uniformly Cauchy. But for distinct indices \(m,n\), \[ \norm{f_n-f_m}_\infty = \abs{n-m} \geq1. \] Hence no subsequence is uniformly Cauchy.
The graph of \(f_n\) is the horizontal line of height \(n\).
3. Conditions (ii) and (iii) can hold while (i) fails, without precompactness.
Take \[ \Omega=\RR \] and define \[ \phi(t)=\max\{1-2\abs{t},0\}, \qquad f_n(x)=\phi(x-n). \] The domain \(\RR\) is not compact. Each \(f_n\) satisfies \[ 0\leq f_n(x)\leq1, \qquad \norm{f_n}_\infty=1, \] so (ii) holds.
The function \(\phi\) is \(2\)-Lipschitz, hence for every \(n\), \[ \abs{f_n(x)-f_n(y)} \leq 2\abs{x-y}. \] Thus the family is equicontinuous.
If \(m\neq n\) are positive integers, then \[ f_n(n)=1 \] whereas \[ f_m(n)=0, \] because \(\abs{n-m}\geq1\) and \(\phi\) vanishes outside \([-1/2,1/2]\). Consequently \[ \norm{f_n-f_m}_\infty\geq1. \] No subsequence is uniformly Cauchy, hence no subsequence converges uniformly.
The graph of \(f_n\) is a triangular bump of height \(1\) centered at \(x=n\), supported on \[ \left[n-\frac12,n+\frac12\right], \] and zero elsewhere.
4. Q.E.D.
Steps 1, 2, and 3 give the three requested examples and describe their typical graphs.