Lecture 13

Inversion and Duality¶

Recall that for $f \in \mathcal S$ we have the duality formula $\mathcal F(f^-) = (\mathcal Ff)^- = \mathcal F^{-1}f$, which follows directly from the definition of inverse Fourier transform and Fourier transform. This is also the case for tempered distributions.

Let $T$ be a tempered distribution. Let $T^{-}$ be the tempered distribution given by $(T^-,g) = (T,g^{-})$.

For $T\in \mathcal S'$, let $\mathcal F^{-1}T$ be the tempered distribution given by $(\mathcal F^{-1}T,g) = (T,\mathcal F^{-1} g)$. Then $(\mathcal F^{-1}\mathcal F T,g) = (\mathcal FT,\mathcal F^{-1}g) = (T,\mathcal F \mathcal F^{-1}g) = (T,g)$. Then we obtain

Then $(\mathcal FT^{-},g) = (T^-,\mathcal Fg) = (T,(\mathcal Fg)^{-}) = (T,\mathcal F^{-1}g) = (T,\mathcal Fg^{-})$.

Remark on the duality formula¶

We developed the inversion theory as follows:

• First define inverse Fourier transform $\mathcal F^{-1}f$ by $\mathcal F^{-1}f(s) = \int_{-\infty}^{+\infty}f(x)e^{2\pi i xs}dx$.
• Use the periodization lemma to prove that $f(x) = \int_{-\infty}^{+\infty}\mathcal Ff(s)e^{2\pi i x s}ds$, then conclude the inversion theorem $\mathcal F^{-1}\mathcal F f = f$.
• The duality formula $\mathcal F\mathcal F^- = \mathcal F^{-1}$ is a consequence of inversion theorem.

In fact, there is an alternative approach, see problem set 6.

• First define the inversion operator $\mathcal F^{-1}$ to be $\mathcal F^{-1}f = \mathcal Ff ^-$.
• Let $Tf = \mathcal F^{-1}\mathcal F f = \mathcal F\mathcal F f^-$. Observe that $T$ commutes with multiplication by $x$ and $\frac{d}{dx}$.
• Conclude that $T$ must be the identity!
• This implies the inversion theorem!

Convolutions¶

Convolution of Schwartz function with tempered distribution¶

We aim to define the convolution $f*T$ for $f\in \mathcal S, T\in \mathcal S'$.

By the “try function first” principle, consider the case $T = T_g$ for smooth function $g$. Then

Alternatively, instead of using $f^-$, we can also change variable $x' = x-y$ to rewrite the integral as $\int_{y}g(y)\int_x f(x')h(x'+y)dx' = (T_g(y),(T_f(x),h(x+y))$, where we treat $(T_f(x),h(x+y)) = (T,\tau_{-y}h)$ as a function of $y$.

Shifting and Scaling¶

Recall for $f\in \mathcal S$, we have proved the shifting theorem $\mathcal F(f(x-a))(s) = e^{-2\pi i a s}\mathcal Ff(s)$ and the scaling theorem $\mathcal Ff(ax)(s) = \frac{1}{|a|}\mathcal Ff(\frac{s}{a})$. In order to formulate the correspondent properties for distributions, we introduce the shifting operator $\tau_a:\mathcal S\to \mathcal S$, $\tau_af (x) = f(x-a)$, and the scaling operator $\sigma_a:\mathcal S\to\mathcal S$, $\sigma_a f(x) = f(ax)$. Then the shifting theorem and scaling theorem is given by

Shifting of tempered distribution¶

Using the “try function” principle, $(\tau_a T_f,g) = \int f(x-a)g(x)dx = \int f(x')g(x'+a)dx' = (T_f,\tau_{-a}g)$. Then we define $\tau_a T$ by $(\tau_a T,g) = (T,\tau_{-a} g)$.

Scaling of tempared distribution¶

$(\sigma_a T_f,g) = \int f(ax)g(x)dx = \frac{1}{|a|}\int f(x')g(\frac{x'}{a})dx' = (T_f,\frac{1}{|a|}\sigma_{\frac{1}{a}}g)$. Then we define $\sigma_a T$ by $(\sigma_a T,g) = (T,\frac{1}{|a|}\sigma_{\frac{1}{a}}g)$.

Shifting and scaling theorem¶

$(\mathcal F\tau_aT,g) = (\tau_a T, \mathcal F g) = (T,\tau_{-a}\mathcal F g) = (T,\mathcal F(e^{-2\pi i a x}\cdot g)) = (\mathcal FT,e^{-2\pi i a x}\cdot g) = (e^{-2\pi i a x }\mathcal FT, g)$.

$(\mathcal F\sigma_aT,g) = (\sigma_aT, \mathcal F g) = (T,\frac{1}{|a|}\sigma_{\frac{1}{a}}\mathcal Fg) = (T,\frac{1}{|a|}\cdot |a|\mathcal F\sigma_{a}g) = (T,\mathcal F\sigma_a g) = (\mathcal FT,\sigma_a g) = (\frac{1}{|a|}\sigma_a\mathcal FT,g)$.

From the above derivation we get

It’s formally the same as the case of $\mathcal S$.