Find the pointwise limit of the sequence of functions \(\{f_n\}\) given by \(f_n(x)=x^n\) on \([0,1]\). Is the convergence of \(f_n\) to \(f\) uniform? ([KRD10, 8.6.A]) Why is \[B=\{f\in C([0,1]):\|f\|_\infty\le1\}\] not compact?
Solution
Goal: (1) Find the pointwise limit of \(f_n(x) = x^n\) on \([0,1]\); (2) decide uniformity; (3) explain why \(B = \{f \in C[0,1] : \|f\|_\infty \le 1\}\) is not compact.
1. The pointwise limit is \(f(x) = 0\) for \(0 \le x < 1\) and \(f(1) = 1\).Proof: for \(0 \le x < 1\), \(x^n \to 0\); at \(x = 1\), \(x^n = 1\) for all \(n\).
2. The convergence is not uniform on \([0,1]\).
2.1. Each \(f_n\) is continuous.
2.2. \(f\) is discontinuous at \(1\): \(\lim_{x \to 1^-} f(x) = 0 \neq 1 = f(1)\).Proof: by
1.
1.3. If \(f_n \to f\) uniformly, \(f\) would be continuous.Proof: uniform limits of continuous functions are continuous (e.g. Theorem 6.1 / D7-W3).
1.4. Q.E.D. Proof: 2 contradicts
1.3. Alternatively: \(\|f_n - f\|_\infty = \sup_{x \in [0,1)}x^n = 1 \not\to 0\) — the sup is approached as \(x \to 1^-\).
3. \(B = \{f \in C[0,1] : \|f\|_\infty \le 1\}\) is not compact.
3.1. \(f_n(x) = x^n\) is a sequence in \(B\) with no uniformly convergent subsequence.Proof: every subsequence \(f_{n_k}\) converges pointwise to the same discontinuous \(f\) of 1; if some subsequence converged uniformly, its uniform limit (necessarily \(f\), by uniqueness of pointwise limits) would be continuous, contradiction.
3.2. In a compact metric space every sequence has a convergent subsequence.Proof: sequential compactness, equivalent to compactness for metric spaces.
3.3. Q.E.D. Proof: 1 gives a sequence in \(B\) without a convergent subsequence, so \(B\) is not compact by
3.2. (Intuitively: \(B\) is closed and bounded but not equicontinuous, so Arzelà–Ascoli fails.)