Problem of the Week #32 - November 5th, 2012

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Chris L T521

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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Let $V$ be the space of differentiable complex-valued functions on the unit circle in the complex plane, and for each $f,g\in V$, define
$\langle f,g\rangle=\int_0^{2\pi}\overline{f(\theta)} g(\theta) \,d\theta.$
Show that this form is Hermitian (i.e. $\langle f,g\rangle = \overline{\langle g,f\rangle}$) and positive definite (i.e. $\langle f,f\rangle > 0$ for all nonzero functions $f\in V$).

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Chris L T521

Well-known member
Staff member
This week's question was correctly answered by Sudharaka. You can find his solution below.

\begin{eqnarray}

\langle f,g\rangle&=&\int_0^{2\pi}\overline{f(\theta)} g(\theta) \,d\theta\\

&=&\int_0^{2\pi}\overline{f(\theta) \overline{g(\theta)}} \,d\theta\\

\end{eqnarray}

For a complex valued function of a real variable, $$\int_{a}^{b}\overline{f(x)}\,dx=\overline{\int_{a}^{b}f(x)\,dx}$$. Therefore,

\begin{eqnarray}

\langle f,g\rangle&=&\overline{\int_0^{2\pi}\overline{g( \theta)}f(\theta) \,d\theta}\\

&=&\overline{\langle g,f\rangle}

\end{eqnarray}

\begin{eqnarray}

\langle f,f\rangle&=&\int_0^{2\pi}\overline{f( \theta)}f(\theta) \,d\theta\\

\end{eqnarray}

Let, $$f(\theta)=f_{1}(\theta)+if_{2}(\theta)$$ where $$f_{1}$$ and $$f_{2}$$ are real valued functions. Then we get,

$\langle f,f\rangle=\int_0^{2\pi}[f_{1}(\theta)]^{2} \,d\theta+\int_0^{2\pi}[f_{2}( \theta)]^{2}\,d\theta$

$\therefore\langle f,f\rangle>0\mbox{ for any non zero function }f\in V$

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