MHB Problem of the Week #116 - August 18th, 2014

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The problem discusses the vector space of smooth functions, known as the Schwarz space, and defines a function \( P \) that integrates the product of two functions from this space. It confirms that \( P \) defines a pairing by demonstrating linearity in both arguments. Ackbach provided a partial solution, establishing that \( P \) satisfies the necessary conditions for a pairing. However, the non-degeneracy of \( P \) remains unresolved, with further exploration expected in the coming days. The discussion highlights the ongoing analysis of properties related to smooth functions and their integrals.
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Here's this week's problem!

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Problem
: Let $V$ be the vector space of smooth (i.e. infinitely-differentiable) functions $f:\mathbb{R}\rightarrow\mathbb{C}$ such that for all $m,n\in\mathbb{N}$, there exists $C_{m,n}>0$ for which$$\forall\,x\in\mathbb{R},\,(1+x^2)^m\cdot|f^{(n)}(x)|\leq C_{m,n}.$$
This is called the Schwarz space - its elements are smooth functions all of whose derivatives decay rapidly. Define a function $P:V\times V\rightarrow\mathbb{C}$ by
$$P(f,g)=\int_{-\infty}^{\infty}f(x)g(x)\,dx.$$
Does $P$ define a pairing? Is $P$ nondegenerate?

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This week's delays are brought to you by yours truly starting work again now that Fall semester is underway.

This week's problem was partially correctly answered by Ackbach. As for the case of whether or not the pairing is non-degenerate, I too am still working on that part and I hope to have a solution posted sometime in the next couple days.

Without much further ado, you can find Ackbach's partial solution below.

[sp]$P$ definitely defines a pairing. Check:
$\begin{align*}
P(r f,g)&=\int_{-\infty}^{\infty}(rf(x))g(x) \, dx \\
&=\int_{-\infty}^{\infty}f(x)(rg(x)) \, dx \\
&=P(f,rg) \\
&=r \int_{-\infty}^{\infty}f(x) g(x) \, dx \\
&=r P(f,g).
\end{align*}$
Moreover,
$\begin{align*}
P(f+g,h)&=\int_{-\infty}^{\infty}(f(x)+g(x))h(x) \, dx \\
&=\int_{-\infty}^{\infty}f(x) h(x) \, dx+\int_{-\infty}^{\infty}g(x) h(x) \, dx \\
&=P(f,h)+P(g,h),
\end{align*}$
and similarly, $P(f,g+h)=P(f,g)+P(f,h)$. Thus, $P$ satisfies all the requirements for a pairing.[/sp]