Let \(f_n(x) = n^\beta x(1-x^2)^n\), \(x \in [0,1]\), \(n \in \mathbb{N}\).
Prove that \(\{f_n\}_{n=1}^\infty\) converges pointwise on \([0,1]\) for every \(\beta \in \mathbb{R}\).
Show that the convergence in part (a) is uniform for all \(\beta < \frac{1}{2}\), but not uniform for any \(\beta \geq \frac{1}{2}\).
Solution
1. \(f_n \to 0\) pointwise on \([0,1]\) for every \(\beta \in \RR\).
1.1. For \(x = 0\): \(f_n(0) = 0\) for all \(n\).
the factor \(x\) vanishes.
1.2. For \(0 < x \le 1\): \(f_n(x) = n^\beta x(1 - x^2)^n \to 0\).
\((1 - x^2)^n\) decays exponentially in \(n\) for \(x > 0\) (\(1 - x^2 \in [0,1)\)), which beats the polynomial growth \(n^\beta\); more precisely \(n^\beta(1-x^2)^n = n^\beta e^{n\ln(1-x^2)} \to 0\) since \(\ln(1 - x^2) < 0\).
1.3. Q.E.D.
1 and 2.
2. \(\sup_{x \in [0,1]}|f_n(x)| = n^\beta \cdot \frac{1}{\sqrt{2n+1}}\left(\frac{2n}{2n+1}\right)^n\).
2.1. Maximize \(g(x) = x(1 - x^2)^n\): \(g'(x) = (1 - x^2)^{n-1}(1 - (2n+1)x^2)\), so the maximum on \([0,1]\) is at \(x_n = 1/\sqrt{2n+1}\).
derivative computation; \(g' > 0\) for \(x < x_n\) and \(g' < 0\) for \(x > x_n\); \(g(0) = g(1) = 0\).
2.2. \(g(x_n) = \frac{1}{\sqrt{2n+1}}\left(1 - \frac{1}{2n+1}\right)^n = \frac{1}{\sqrt{2n+1}}\left(\frac{2n}{2n+1}\right)^n\).
\(1 - x_n^2 = 1 - \frac{1}{2n+1} = \frac{2n}{2n+1}\).
2.3. Q.E.D.
\(\|f_n\|_\infty = n^\beta g(x_n)\).
3. Uniform convergence holds iff \(\beta < \frac{1}{2}\).
3.1. \(\|f_n\|_\infty \asymp n^{\beta - 1/2}\): \(\|f_n\|_\infty = n^\beta \cdot \frac{1}{\sqrt{2n+1}}\left(\frac{2n}{2n+1}\right)^n\), and \(\left(\frac{2n}{2n+1}\right)^n = \left(1 - \frac{1}{2n+1}\right)^n \to e^{-1/2} > 0\).
\(\frac{1}{\sqrt{2n+1}} \sim \frac{1}{\sqrt 2}n^{-1/2}\); the factor \(\left(\frac{2n}{2n+1}\right)^n\) has a positive limit \(e^{-1/2}\).
3.2. If \(\beta < 1/2\): \(\|f_n\|_\infty \to 0\), so \(f_n \to 0\) uniformly.
\(n^{\beta - 1/2} \to 0\).
3.3. If \(\beta \ge 1/2\): \(\|f_n\|_\infty \not\to 0\), so the convergence is not uniform.
\(n^{\beta - 1/2} \not\to 0\) (it diverges for \(\beta > 1/2\) and is bounded away from \(0\) for \(\beta = 1/2\)), and uniform convergence to the pointwise limit \(0\) would force \(\|f_n\|_\infty \to 0\).
3.4. Q.E.D.
2 and 3.