Question 1.3. Let \(f _ { i } : [ 0 , 1 ] \to \mathbb { R } , i = 1 , 2 , . . . ,\) , be an increasing sequence of continuous functions that is uniformly bounded, i.e., for all \(x \in [ 0 , 1 ]\) and \(i \geq 1 , f _ { i } ( x ) \leq f _ { i + 1 } ( x ) \leq\) \(M < \infty f o r\) some fixed M. Show that li \(\mathfrak { a } _ { i \to \infty } f _ { i } ( x ) = g ( x )\) is continuous if and only \(i f\) the \(f _ { i }\) converge uniformly to g.
Solution
1. The pointwise limit exists.
For each fixed \(x\in[0,1]\), the sequence \((f_i(x))\) is increasing and bounded above by \(M\). Hence \[ g(x):=\lim_{i\to\infty}f_i(x) \] exists for every \(x\in[0,1]\).
2. Uniform convergence implies continuity of the limit.
If \(f_i\to g\) uniformly on \([0,1]\), then \(g\) is continuous because a uniform limit of continuous functions is continuous.
3. Continuity of the limit implies uniform convergence.
Assume now that \(g\) is continuous. Define \[ h_i:=g-f_i. \] Then each \(h_i\) is continuous, \(h_i\ge0\), and \[ h_1\ge h_2\ge\cdots\downarrow0 \] pointwise.
Fix \(\varepsilon>0\) and set \[ U_i:=\{x\in[0,1]:h_i(x)<\varepsilon\}. \] Each \(U_i\) is open in \([0,1]\), the sequence \((U_i)\) is increasing, and pointwise convergence gives \[ \bigcup_{i=1}^\infty U_i=[0,1]. \] Because \([0,1]\) is compact and the \(U_i\) are increasing, there exists \(N\) such that \[ [0,1]\subseteq U_N. \] Hence for every \(x\in[0,1]\), \[ 0\le g(x)-f_N(x)<\varepsilon. \] Since \(f_i\ge f_N\) for \(i\ge N\), \[ 0\le g(x)-f_i(x)\le g(x)-f_N(x)<\varepsilon \] for all \(x\in[0,1]\) and all \(i\ge N\). Therefore \[ \|g-f_i\|_\infty<\varepsilon \] for every \(i\ge N\).
Thus \(f_i\to g\) uniformly. Equivalently, for this monotone sequence of continuous functions on the compact interval, the pointwise limit is continuous if and only if the convergence is uniform.