Exercise 11.13

Question

Consider the functions
    $$ f_n(x) = \sin nx \quad (n = 1,2,3,\ldots,\ -\pi \le x \le \pi) $$
    as points of $\mathscr{L}^2$. Prove that the set of these points is closed
    and bounded, but not compact.

Amit Awasthi · Accepted Answer

\begin{proof}
      We have
      \begin{align*}
        \lVert f_n Vert = \int_{-\pi}^\pi \sin^2(nx)\,dx
        &= \frac{1}{n} \int_{-n\pi}^{n\pi} \sin^2(u)\,du\
        &= \int_{-\pi}^\pi \sin^2(u)\,du = \pi.
      \end{align*}
      Moreover,
      \begin{align*}
        &\lVert f_n - f_m Vert\
        &\quad= \int_{-\pi}^\pi (f_n - f_m)^2\,dx\
        &\quad= \lVert f_n Vert + \lVert f_m Vert - 2\int_{-\pi}^\pi
        \sin(nx)\sin(mx)\,dx\
        &\quad= 2\pi (1-\delta_{nm}),
      \end{align*}
      so the set of $f_n$'s is bounded. Since every point is isolated, it
      contains all limit points, hence the set is also closed.
      The set of $f_n$'s is not compact since the cover consisting of balls
      of radius $\pi$ arond each $f_n$ does not have a finite subcover.
    \end{proof}

Exercise 11.13

Answers

Comments

Exercise 11.13

Answers

Comments

Add answer