- Prove the following partial converse to the uniform limit theorem.
Theorem. Let \(f_n: X \to \mathbb{R}\) be a sequence of continuous functions, with \(f_n(x) \to f(x)\) for each \(x \in X\). If \(f\) is continuous, and if the sequence \(f_n\) is monotone increasing, and if \(X\) is compact, then the convergence is uniform. [We say that \(f_n\) is monotone increasing if \(f_n(x) \leq f_{n+1}(x)\) for all \(n\) and \(x\).]
- Give examples to show that this theorem fails if you delete the requirement that \(X\) be compact, or if you delete the requirement that the sequence be monotone. [Hint: See the exercises of §21.]
Solution
This is Dini’s theorem. Fix \(\varepsilon>0\). Since \(f_n\uparrow f\), define \[ U_n=\{x\in X:f(x)-f_n(x)<\varepsilon\}. \] Because \(f-f_n\) is continuous, \(U_n\) is open. The sets are increasing, since \(f_n\le f_{n+1}\), and pointwise convergence implies \(\bigcup_nU_n=X\). Compactness gives a finite subcover; since the family is increasing, there is some \(N\) with \(U_N=X\). Thus for all \(x\in X\), \[ 0\le f(x)-f_N(x)<\varepsilon. \] For every \(n\ge N\), monotonicity gives \[ 0\le f(x)-f_n(x)\le f(x)-f_N(x)<\varepsilon, \] uniformly in \(x\). Hence \(f_n\to f\) uniformly.
Compactness is necessary. On \(X=\mathbb R\), let \[ f_n(x)=\frac{n}{n+|x|}. \] Each \(f_n\) is continuous, \(f_n(x)\uparrow 1\) for each fixed \(x\), but \[ \sup_x |1-f_n(x)|=1, \] so convergence is not uniform.
Monotonicity is also necessary even on a compact domain. On \([0,1]\), let \[ f_n(x)=x^n(1-x). \] Then \(f_n\to0\) pointwise and the limit is continuous, but this particular sequence actually converges uniformly, so instead use the standard moving-spike example: choose continuous triangular functions \(g_n:[0,1]\to[0,1]\) supported in \((\frac1{n+1},\frac1{n-1})\) with \(g_n(1/n)=1\). Then \(g_n(x)\to0\) for every fixed \(x\), because each \(x>0\) lies in only finitely many supports and \(g_n(0)=0\); but \(\sup g_n=1\) for all \(n\). Thus convergence is not uniform.