Stepan Malkov

Week 7-8 Problems

Easy:

  1. Find the radius of convergence of the power series

        \[\sum_{n=1}^\infty \frac{n(x+2)^n}{5^{n-1}}.\]

    2. Proof:
    By using the root test, we obtain

        \[\limsup_{n \to \infty} |a_n|^{\frac{1} {n}} = \limsup_{n \to \infty} \frac{n^{\frac{1}{n}}}{5^{\frac{n-1}{n}}} = \frac{1}{5},\]

    so R = 5.
  2. Provide an example of a real power series \sum_{n=0}^\infty a_n x^n with radius of convergence R that converges at x=-R, but diverges at the endpoint x=R.
    3. Proof:
    Take f(x) = \ln(1-x) = \sum_{n=1}^\infty \frac{x^n}{n}. With radius of convergence R=1. Then, f(1) = \sum_{n=0}^\infty \frac{1}{n} = \infty is the well-known harmonic series, which diverges, but

        \[f(-1) = \ln(2) = \sum_{n=1}^\infty \frac{(-1)^n}{n}\]

    converges by the Alternating Series Test.
  3. Let X=C(\mathbb{R}/\mathbb{Z},\mathbb{R}) be the space of continuous periodic functions on [0,1]. Define the L^2 norm on X to be

        \[\|f\|_{L^2} = \left(\int_0^1 |f(x)|^2 dx\right)^{\frac{1}{2}}.\]

    Show that this is indeed a norm on X. Conclude that the L^2 metric on X is

        \[d(f,g)_{L^2}= \left(\int_0^1|f(x)-g(x)|^2dx \right)^{\frac{1}{2}}.\]

    4. Proof:
    Indeed \|0\|_{L^2}=0, \|c f\|_{L^2}=|c|\|f\|_{L^2}, and if f \not = 0, there exist x such that f(x) \not =0, and by continuity, there exists a \delta>0 such that |f(x)-f(y)|<\frac{|f(x)|}{2} for y \in [x-\delta,x+\delta]. Then,

        \[\|f\|_{L^2} = \left(\int_0^1 |f(x)|^2 dx \right)^{\frac{1}{2}} \geq \left(\int_{x-\delta}^{x+\delta} |f(x)|^2 dx\right)^{\frac{1}{2}} \geq \left(\frac{|f(x)|}{2}\right)^2 (2\delta) > 0,\]

    so \|f\|_{L^2} = 0 if and only if f =0. Finally, for the triangle inequality, note that by Cauchy-Schwarz

        \[\int_0^1|f(x)g(x)| \leq \sqrt{\int_0^1|f(x)|^2dx}\sqrt{\int_0^1 |g(x)|^2dx},\]

    so

        \begin{equation*} \begin{split} \int_0^1|f(x)+g(x)|^2 dx & = \int_0^1 f(x)^2+g(x)^2 + 2|f(x)g(x)|dx \\ & \leq \int_0^1 f(x)^2 dx +2 \sqrt{\int_0^1|f(x)|^2dx}\sqrt{\int_0^1 |g(x)|^2dx} + \int_0^1 g(x)^2 dx \\ & \leq \left(\sqrt{\int_0^1 f(x)^2 dx} + \sqrt{\int_0^1 g(x)^2 dx}\right)^2, \end{split} \end{equation*}

    and taking square roots on both sides yields

        \[\|f+g\|_{L^2} \leq \|f\|_{L^2} + \|g\|_{L^2}.\]

    Thus, \|\cdot\|_{L^2} is indeed a norm on X=C(\mathbb{R}/\mathbb{Z},\mathbb{R})), turning X into a metric space with d_{L^2}(f,g)=\|f-g\|_{L^2}.
  4. Show that if f_n converges to f uniformly, f_n converges to f in L^2.
    5. Proof:
    Indeed, whenever \|f_n-f\|_{\infty}<\epsilon,

        \[\|f_n-f\|_{L^2} = \left(\int_{0}^1 |f_n-f|^2\right)^{\frac{1}{2}} \leq \epsilon,\]

    so uniform convergence implies convergence in L^2.

Medium:

  1. If (x_n)_{n=1}^\infty is a sequence of real numbers, \lim_{n \to \infty} x_n exists if and only if \limsup_{n \to \infty} x_n = \liminf_{n \to \infty} x_n, in which case

        \[\limsup_{n \to \infty} x_n = \liminf_{n \to \infty} x_n = \lim_{n \to \infty} x_n.\]

    2. Proof:
    If \lim_{n \to \infty} x_n = L, then for any \epsilon>0, there is an N such that n \geq N, |x_n - L| <\epsilon, i.e. |\sup_{n \geq N} a_n - L| \leq \epsilon, so \limsup_{n \to \infty} a_n = L. Similarly \liminf_{x_n \to \infty} x_n = L. Conversely, suppose \limsup_{x_n \to \infty} = \liminf_{x_n \to \infty} x_n = L. Then, for any \epsilon > 0, there exists N = \max(N_1,N_2), where |\sup_{n \geq N_1} a_n - L| <\epsilon, |\sup_{n \geq N_2} a_n - L| < \epsilon, i.e.

        \[|a_n-L|<\epsilon\]

    for n \geq N, implying that

        \[\lim_{n \to \infty} x_n= \limsup_{n \to \infty} x_n = \liminf_{n \to \infty} x_n = L.\]

  2. (Ratio Test) The radius of convergence of \sum_{n=1}^\infty a_n z^n is \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right| if the value of the limit exists.
    3. Proof:
    Note that by the ratio test for series, the series above converges whenever \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n} z\right|<1, i.e. whenever |z|< \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|=R, and will diverge otherwise. Thus, by definition, R is the radius of convergence of the power series.
  3. Show that if

        \[f(x) = \sum_{n=0}^\infty a_n x^n\]

    is a real power series with radius of convergence R, then g(y) = \int_{0}^y f(x)dx is differentiable on (-r,r) and g'(y) = f(y) for all y \in (-R,R). Moreover, show that

        \[g(y) = \sum_{n=0}^\infty \frac{a_n}{n+1} y^{n+1}.\]

    4. Proof:
    Note that since f is continous for any \epsilon>0, there exists an h such that |f(y+h)-f(y)|<\epsilon for h<\delta. But for such an h,

        \[\frac{g(y+h)-g(y)}{h} - f(y) = \frac{\int_0^{y+h} f(x) dx - \int_0^y f(x) dx} {h} - f(y) = \frac{1}{h}\int_y^{y+h} f(x) - f(y)dx \leq \frac{1}{h} \epsilon = \epsilon,\]

    i.e. \lim_{h \to 0} \frac{g(y+h)-g(y)}{h} = f(y), so g is differentiable on (-R,R) and g'(y) = f(y). Then, since the sum convergees uniformly to f, by the interchangeability of the integral and limit, one gets

        \[g(y) = \int_0^y \sum_{n=0}^\infty a_n x^n dx = \sum_{n=0}^\infty \int_0^y a_n x^n dx = \sum_{n=0}^\infty \frac{a_n}{n+1} y^{n+1}.\]

Challenging:

  1. Find the Fourier transform of \sin 2\pi x.
    2. Proof:
    Using the identity

        \[\sin x =\frac{e^{ix}-e^{-ix}}{2i},\]

    one obtains

        \begin{equation*} \begin{split} \mathcal{F}(\sin 2 \pi x)(n) & = \int_0^1 \frac{e^{2\pi i x} - e^{-2\pi i x}}{2i} e^{-2\pi i n x} dx = \int_0^1 \frac{e^{2\pi i x(1-n)} -e^{-2\pi i x (n+1)}}{2i} dx \\ & = \left[\frac{e^{2\pi i x(1-n)}}{2i(2\pi i)(1-n)}\right]^1_0-\left[\frac{e^{-2\pi i x (n+1)}}{2i(-2\pi i (n+1))}\right]^1_0 = \frac{e^{2\pi i (1-n)}-1}{4\pi (n-1)}-\frac{e^{-2\pi i (n+1)}-1}{4\pi (n+1)}. \end{split} \end{equation*}

  2. Prove Bessel’s inequality: for f \in C(\mathbb{R}/\mathbb{Z},\mathbb{R}) and any N>0,

        \[\sum_{n=1}^N |\langle f, e_n \rangle|^2 \leq \|f\|^2_{L^2}.\]

    3. Proof:
    Let y = \sum_{n=1}^N \langle f, e_n \rangle e_n. Then, \langle y, e_n \rangle = \langle x,e_n \rangle for all n, i.e. \langle x-y, y\rangle =0. Then, since y is perpendicular to x-y, by the Pythagorean theorem,

        \[\|x\|^2= \|x-y\|^2 + \|y\|^2 \geq \|y\|^2= \sum_{n=1}^N |\langle f, e_n \rangle|^2.\]