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This is a problem I found in Schaum's Outlines: Linear Algebra, and I was wondering if someone knew how to solve it. I began using integration by parts, but that approach did not lead to any conclusions.

Let V be the space of all infinitely-differentiable functions on R which are periodic of period h>0 [i.e., f(x+h) = f(x) for all x in R]. Define an inner product on V by $$\langle f,g\rangle =\int_{-h}^hf(x)g(x)dx$$ Let $\alpha(f)=f'$. Find $\alpha^*$.

I know that the adjoint implies the relationship $\langle\alpha(f),g\rangle= \langle f,\alpha^*(g)\rangle$ .

Thank you.

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  • $\begingroup$ Could you define $\alpha^{*}$? $\endgroup$
    – Jebruho
    Nov 4, 2012 at 22:29
  • $\begingroup$ Here is a related problem. $\endgroup$
    – Mhenni Benghorbal
    Nov 4, 2012 at 22:36
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    $\begingroup$ This result is very interesting -- a special case of the fact that the adjoint of $\nabla$ is $-\text{div}$. It also implies that $\frac{d}{dt}$ is a "normal" operator, so one would hope that the spectral theorem should apply: there should be an orthonormal basis of eigenvectors of $\frac{d}{dt}$. This idea leads us to discover Fourier series. $\endgroup$
    – littleO
    Nov 4, 2012 at 22:51

2 Answers 2

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We have $$\langle \alpha(f),g\rangle=\int_{-h}^hf'(t)g(t)dt=\left[f(t)g(t)\right]_{-h}^h-\int_{-h}^hf(t)g'(t)dt.$$ As $f\cdot g$ is periodic of period $h$, $\left[f(t)g(t)\right]_{-h}^h=0$, so $\alpha^*(f)=-\alpha(f)$.

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Yes, $\left.f(x)g(x)\right|_{-h}^{+h}=0$ because $f(x+h)=f(x) \forall x \in \Re$, i.e. with $x=0$, $f(h)=f(0)$; and with $x=-h$, $f(0)=f(-h)$. The same occurs with $g(x)$, so $f(h)g(h)=f(0)g(0)=f(-h)g(-h)$, and $\left.f(x)g(x)\right|_{-h}^{+h}=0$. QED.

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