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Set-Up

Let $f$ be a tempered distribution given by integration against a function (which we also call $f$). Suppose that there is a function $g \in L^2$ such that $\mathcal{F}f = g$. In other words, the tempered distribution $\mathcal{F}f$ is given by integration against an $L^2$ function.

Question

Do we have $$ \int |f|^2 = \int |g|^2 ? $$

Observations

  • $\mathcal{F}^{-1}(g)$ is an $L^2$ function, so it suffices to prove $f(x)=\mathcal{F}^{-1}(g)(x)$ as functions for a.e. $x$.

  • $\int f \psi = \int \mathcal{F}^{-1}(g) \psi$ for all Schwartz functions $\psi$, so that $f = \mathcal{F}^{-1}(g)$ as tempered distributions.

  • If $f$ and $g$ are continuous bounded functions, then $$ f(x) = \lim_{\epsilon \to 0} f \ast \psi_{\epsilon} (x) = \lim_{\epsilon \to 0} g \ast \psi_{\epsilon} (x) = g(x) \quad \text{for all $x$}, $$ where $\psi_{\epsilon}$ is a Schwartz approximation to the identity.

Sub-Question If two functions are equal as tempered distributions, under what conditions are they equal (or equal a.e.) as functions?

Special Case

This special case is connected to Convolution theorem with distributions

  • Take $f = u \ast \varphi$, where $\varphi \in S$ (Schwartz space) and $u \in S'$ (space of tempered distributions).

  • With appropriate conditions on $u$, the tempered distribution $\mathcal{F}(u \ast \varphi)$ is integration against a function $g \in L^2$.

  • Note $u \ast \varphi$ is the function defined by $$ (u \ast \varphi)(x) = \langle u, R \tau_x \varphi \rangle, $$ where $\tau_x \phi(y)=\phi(y-x)$ and $R\tau_x\phi(y) = \phi(x-y)$. In the particular case where $u$ is a function, $$ (u \ast \varphi)(x) = \int u(y) \phi(x-y) dx. $$ Note also $u \ast \varphi$ is a $C^{\infty}$ function and a tempered distribution.

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For the first question, note that \begin{align*} \left<\mathcal{F}^{-1}(\mathcal{F}L_{f}),h\right>&=\left<\mathcal{F}L_{f},\mathcal{F}^{-1}(h)\right>\\ &=\left<L_{f},\mathcal{F}\mathcal{F}^{-1}h\right>\\ &=\left<L_{f},h\right>\\ &=\left<f,h\right>, \end{align*} where $h\in C_{0}^{\infty}({\bf{R}}^{n})$, so $\mathcal{F}^{-1}\mathcal{F}L_{f}=f$ a.e., that is, $\mathcal{F}^{-1}g=f$ a.e. Since $\mathcal{F}$ is a bijection in $L^{2}$, so $f\in L^{2}$ and that $\|f\|_{L^{2}}=\|\mathcal{F}(f)\|_{L^{2}}=\|g\|_{L^{2}}$ follows by the isometric property of $\mathcal{F}$, which is Parseval's formula.

For the second question, it must be the case that $f=g$ a.e. Actually a theorem says that if \begin{align*} \int_{{\bf{R}}^{n}}f(x)h(x)=\int_{{\bf{R}}^{n}}g(x)h(x)dx,~~~~h\in C_{0}^{\infty}({\bf{R}}^{n}), \end{align*} then $f=g$ a.e., where $C_{0}^{\infty}({\bf{R}}^{n})$ means all $C^{\infty}({\bf{R}}^{n})$ functions that having compactly supports and here $f,g\in L_{\text{loc}}^{1}({\bf{R}}^{n})$. The same result is true if we consider any open set $\Omega$ instead of ${\bf{R}}^{n}$.

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  • $\begingroup$ Is your second assertion really true for arbitrary $f,g$? And do I presume correctly that $C_0^{\infty}$ means $C^{\infty}$ with compact support (rather than vanishing at infinity)? $\endgroup$ – LucasSilva Mar 11 '18 at 17:39
  • $\begingroup$ $f,g\in L_{\text{loc}}^{1}({\bf{R}}^{n})$. $\endgroup$ – user284331 Mar 11 '18 at 17:41

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