Let \[f_n(x)=\frac{1}{1+n^2x^2}\qquad\text{and}\qquad g_n(x)=nx(1-x)^n,\qquad x\in[0,1].\] Prove that \(\{f_n\}\) and \(\{g_n\}\) converge pointwise but not uniformly on \([0,1]\).
Are the families \(\{f_n\}\), respectively \(\{g_n\}\) given in part (a) equicontinuous? Clearly motivate your answer.
Solution
Goal: (a) Prove \(f_n(x) = \frac{1}{1+n^2x^2}\) and \(g_n(x) = nx(1-x)^n\) converge pointwise but not uniformly on \([0,1]\); (b) decide equicontinuity of the two families.
1. (a) \(f_n\) converges pointwise to \(f(x) = 1\) at \(x = 0\) and \(f(x) = 0\) on \((0,1]\).Proof: \(f_n(0) = 1\) for all \(n\); for \(x > 0\), \(n^2x^2 \to \infty\), so \(f_n(x) \to 0\).
2. \(f_n\) does not converge uniformly on \([0,1]\).Proof: the pointwise limit \(f\) is discontinuous at \(0\), while each \(f_n\) is continuous; a uniform limit of continuous functions is continuous.Equivalently \(\|f_n - f\|_\infty = 1\) for all \(n\) (as \(x \to 0^+\), \(f_n(x) \to 1\) while \(f(0) = 1\) and \(f_n(0) = 1\): the sup of \(|f_n - f|\) is approached, e.g. \(\|f_n - f\|_\infty \ge \lim_{x \to 0^+}|f_n(x) - f(x)|\)).
3. (a) \(g_n\) converges pointwise to \(g \equiv 0\).Proof: \(g_n(1) = 0\) and \(g_n(0) = 0\); for \(x \in (0,1)\), \(nx(1-x)^n \to 0\) since exponential decay beats linear growth.
4. \(g_n\) does not converge uniformly.
4.1. \(g_n\) attains its maximum at \(x_n = \frac{1}{n+1}\) with \(g_n(x_n) = \frac{n}{n+1}\left(1 - \frac{1}{n+1}\right)^n \to e^{-1} \neq 0\).Proof: \(\frac{d}{dx}nx(1-x)^n = n(1-x)^{n-1}(1 - (n+1)x)\), zero at \(x = 1/(n+1)\); the value tends to \(e^{-1}\).
4.2. Q.E.D. Proof: 1 gives \(\|g_n\|_\infty \to e^{-1} > 0\), so \(g_n \not\to 0\) uniformly.
5. (b) Neither family is equicontinuous on \([0,1]\).
5.1. \(\{f_n\}\) fails equicontinuity at \(0\): \(f_n(1/n) = \frac{1}{1 + n^2/n^2} = \frac12\) while \(f_n(0) = 1\), so \(|f_n(1/n) - f_n(0)| = 1/2\) even though \(1/n \to 0\).Proof: no \(\delta > 0\) can force \(|f_n(x) - f_n(0)| < 1/4\) for all \(n\) simultaneously, since \(x = 1/n < \delta\) eventually but the difference stays \(1/2\).
5.2. \(\{g_n\}\) fails equicontinuity at \(0\): \(g_n(x_n) \to e^{-1} \neq 0\) at \(x_n = 1/(n+1) \to 0\) while \(g_n(0) = 0\).Proof: by <1>4<2>1, \(|g_n(x_n) - g_n(0)| \to e^{-1} > 0\) with \(x_n \to 0\); again no \(\delta\) works for all \(n\).
5.3. Q.E.D. Proof: 1 and 2 show both families fail the equicontinuity definition at \(0\).