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Problem Sheet 4

Question 1

For the following sequences of functions, discuss the convergence of \(\{f_n\}_{n=1}^\infty\) in the pointwise, uniform, \(L^2\) and \(L^1\) senses.

  1. \(f_n: \mathbb{R} \to \mathbb{R}, f_n(x) = e^{-(x - n)^2}\).
  2. \(f_n: [0, 1] \to \mathbb{R}, f_n(x) = \frac{xe^{-x/n}}{n}\).
  3. \(f_n: [0, 2\pi] \to \mathbb{R}, f_n(x) = \sin(x + \frac{1}{n})\).
  4. \(f_n: [-1, 1] \to \mathbb{R}, f_n(x) = \tan^{-1}(nx)\).

Solution

  1. This is a ''travelling hump'' graph, which converges pointwise to 0. However it cannot converge uniformly since if it did, it would converge to the pointwise limit, but \(\lVert f_n \rVert_\infty = 1\) for all \(n\).

    It also does not converge in the \(L^1\) sense since it is not Cauchy. Note that \(\int_\mathbb{R} f_n(x) \, dx = \sqrt{\pi} > 1\), and so there exists \(r > 0\) such that \(\int_{n - r}^{n + r} f_n(x) = \sqrt{\pi} - \frac{1}{2}\). Take any \(m \in \mathbb{Z}^+\) such that \(m > n + 2r\). Then, using the symmetries of \(f_n\) and \(f_m\),

    \[\begin{align*} d_1(f_n, f_m) &= \int_\mathbb{R} |f_n(x) - f_m(x)| \, dx \\ &\geq \int_{n - r}^{n + r} (f_n(x) - f_m(x)) \, dx + \int_{m - r}^{m + r} (f_m(x) - f_n(x)) \, dx \\ &\geq \left(\sqrt{n} - \frac{1}{2}\right) - \frac{1}{2} + \left(\sqrt{\pi} - \frac{1}{2}\right) - \frac{1}{2} \\ &= 2(\sqrt{n} - 1) > 0. \end{align*}\]

  2. Converges uniformly to 0 since

    \[ f_n(x) = \frac{xe^{-x/n}}{n} \leq \frac{xe^0}{n} \leq \frac{1}{n}, \]

    and thus \(\lVert f_n \rVert_\infty \to 0\). Since we are on a compact interval, we also have \(L^1\) and \(L^2\) convergence.

  3. Converges uniformly to \(\sin\) since the inequality

    \[ |\sin(x + \frac{1}{n}) - \sin(x)| = 2|\cos(\frac{2x + n}{2n}) \sin(\frac{1}{2n})| \leq \frac{1}{n} \]

    holds uniformly for \(x \in [0, 2\pi]\). Since we're on \([0, 2\pi]\), we also have \(L^1\) and \(L^2\) convergence. If the domain was \(\mathbb{R}\), then we have uniform convergence but NOT \(L^1\) nor \(L^2\).

  4. Converges pointwise of \(f\) to \(\frac{\pi}{2}\) for \(x > 0\), \(-\frac{\pi}{2}\) for \(x < 0\) and 0 if \(x = 0\). Pointwise limit is not continuous so cannot converge uniformly. There is convergence to the above function in both \(L^1\) and \(L^2\) done by examining the integral.

Question 2

Give examples of the following if they exist:

  1. A sequence that converges in \(\ell^\infty\) but not in \(\ell^2\).
  2. A sequence that converges in \(\ell^2\) but not in \(\ell^1\).
  3. A sequence of functions on \([0, 1]\) that converges in the \(L^1\) sense but not in the \(L^2\) sense.
  4. A sequence of functions on \([0, 1]\) that converges in the \(L^2\) sense but not in the \(L^1\) sense.

Solution

  1. \(\mathbf{x}_n = (1, \frac{1}{\sqrt{2}}, \dots, \frac{1}{\sqrt{n}}, 0, 0, \dots)\). Let \(\mathbf{x} = (1, \frac{1}{\sqrt{2}}, \dots)\). Then \(\lVert \mathbf{x} - \mathbf{x}_n \rVert_\infty = \frac{1}{\sqrt{n}} \to 0\), but in \(\ell^2\), \(\{\mathbf{x}_n\}_{n = 1}^\infty\) is unbounded, and hence not Cauchy.
  2. \(\mathbf{x}_n = (1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n}, 0, 0, \dots)\).

  3. \(f_n = \sqrt{n}\chi_{[0, 1/n]}\):

    \[\begin{align*} d_1(f_n, 0) &= \int_0^{1/n} \sqrt{n} \, dt = \frac{1}{\sqrt{n}} \to 0 \\ d_2(f_n, f_{2n}) &\geq \int_{1 / (2n)}^{1 / n} \sqrt{n}^2 \, dt = \frac{(\sqrt{2} - 1)^2}{2} + \frac{1}{2}. \end{align*}\]

  4. \(L^2\) converges implies \(L^1\) on an interval by Cauchy-Schwartz inequality:

    \[ \left(\int_0^1 |f_n(t) - f(t)| \, dt \right)^2 \leq \left(\int_0^1 |f_n(t) - f(t)|^2 \, dt \right)\left(\int_0^1 1^2 \, dt\right) = \int_0^1 |f_n(t) - f(t)|^2 \, dt. \]

Question 3

Question 4

Question 5

Given an example to show that you can have \(f_n \to f\) uniformly on \([0, \infty)\) but

\[ \int_0^\infty f_n(t) \, dt \not\to \int_0^\infty f(t) \, dt. \]

Solution

\[ f_n(x) = \frac{e^{-nx}}{n} \]

Question 6

Consider the sequence of functions \(f_n: \mathbb{R} \to \mathbb{R}, n = 1, 2, 3, \dots\) defined by

\[ f_n(x) = \frac{x}{2 + \sqrt{n}x^2, x \in \mathbb{R}}. \]
  1. Does this sequence converge pointwise and/or uniformly on \(\mathbb{R}\)?
  2. Does this sequence of derivatives \(f_n'(x)\) converge pointwise and/or uniformly on \(\mathbb{R}\)?

Solution

  1. For fixed \(x \in \mathbb{R}, f_n(x) \to 0\), and so \(f_n\) converges to 0 pointwise. If it converges uniformly, then the limit must also be 0. Each \(f_n\) is differentiable with derivative

    \[f_n'(x) = \frac{2 - \sqrt{n} x^2}{(2 + \sqrt{n} x^2)^2},\]

    and has a stationary point at \(x = \sqrt{2 / \sqrt{n}}\), which is a local max. Then we see that

    \[f_n(\sqrt{2 / \sqrt{n}}) = \frac{\sqrt{2 / \sqrt{n}}}{2 + \sqrt{n}(2 / \sqrt{n})} = \frac{\sqrt{2}}{4n^{1/4}} \to 0.\]

    So \(f_n \to 0\) uniformly.

  2. For fixed \(x \neq 0\), we will have that \(f_n'(x) \to 0\) and \(f_n'(0) = 0.5\) for all \(n\). Each \(f_n'\) is continuous, and thus cannot converge uniformly.

Question 7