Exercise 11.18

Question

Suppose $f \in \mathscr{L}^2(\mu)$, $g \in \mathscr{L}^2(\mu)$. Prove that
    \begin{equation*}
      \left\lvert \int f\overline{g}\,d\mu \right\rvert^2 = \int \lvert f
      \rvert^2\,d\mu\int\lvert g \rvert^2\,d\mu
    \end{equation*}
    if and only if there is a constant $c$ such that $g(x) = cf(x)$ almost
    everywhere.

Amit Awasthi · Accepted Answer

\begin{proof}
      We claim the equality holds if and only if either $g(x) = 0$ for almost
      all $x$, or ther exists a constant $c$ such that $g(x) = cf(x)$.
      The $\Leftarrow$ direction is clear, so we show the converse.
      We have
      \begin{equation*}
        0 \le \int (\lvert f vert + \lambda \lvert g vert )^2\,d\mu = \lVert
        f Vert^2 + 2\lambda\int\lvert fg vert\,d\mu + \lambda^2 \lVert g
        Vert^2.
      \end{equation*}
      Take $\lambda = - \lvert \int f\overline{g}\,d\mu vert/\lVert g Vert^2$.
      Then,
      \begin{equation*}
        \int (f + \lambda g)\overline{g}\,d\mu = \int f\overline{g}\,d\mu +
        \lambda \lVert g Vert^2 = 0,
      \end{equation*}
      so we also know
      \begin{align*}
        \lVert f Vert^2 &= \int \lvert (f+\lambda g) + \lambda
        gvert^2\,d\mu\
        &= \lVert f + \lambda g Vert^2 + \lVert \lambda g Vert^2\
        &= \lVert f + \lambda g Vert^2 + \frac{\left\lvert f \overline{g}\,d\mu
        ightvert}{\lVert g Vert^2}\
        &= \lVert f + \lambda g Vert^2 + \lVert f Vert^2
      \end{align*}
      and this implies by Exercise 11.2 that $f + \lambda g = 0$ almost
      everywhere as claimed.
    \end{proof}

Exercise 11.18

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Exercise 11.18

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