12.4 Fourier Transforms and Distribution

Proof.

1. (1)
   \begin{align*}\ha{f*g}(\xi)=\integral{\R^n}{}\integral{\R^n}{}e^{-2\pi i x\cd\xi}f(x-y)g(y)dydx=\integral{}{}\integral{}{}e^{-2\pi i(x-y)\cd\xi}f(x-y)e^{-2\pi iy\cd \xi}g(y)dxdy=\ha{f}(\xi)\integral{}{}e^{-2\pi i y\cd\xi}g(y)dy=\ha{f}(\xi)\ha{g}(\xi)\end{align*}

   
   

   by Fubini and translation invariance of Lebesgue measure.

2. (2)

   Change of variable

3. (3)

   Let $y=Tx$ then $x=T\inv y$ and $g(x)\define f(T(x))$ so

   \begin{align*}\ha{f\circ T}(\xi)=\integral{\R^n }{}e^{-2\pi i\xi\cd x}g(x)dx=\integral{\R^n }{}e^{-2\pi i\xi\cd T\inv(y)}f(y)\abs{\det T}\inv dy=\abs{\det T}\inv \integral{}{}e^{-2\pi i((T\inv)^*\xi)\cd y}f(y)dy\end{align*}

   
   

$\square$

Proof. See Folland

1. (1)

   Take $\pd^{\be}$ under $\integral{}{}$ which generate the multiplicative term

2. (2)

   Integration by parts with boundary terms vanishing by $f\in\cali{S}$

3. (3)

   Follows from (1) and (2)

4. (4)

   An application of Fubini shows

   \begin{align*}\integral{}{}f(x)\ha{g}(x)dx=\integral{}{}f(x)\integral{}{}e^{-2\pi ix\cd y}g(y)dydx=\integral{}{}g(y)\integral{}{}e^{-2\pi ix\cd y}f(x)dxdy=\integral{}{}\ha{f}gdx\end{align*}

   
   

$\square$

Proof. We scale $a$ to $1$ by $x\mto a^{-1/2}x$ then

where the last index we use complex analysis. First drop $j$ and let $z=x+i\xi$ since $e^{-\pi z^2}$ is entire and decay rapidly in modulus, for $\abs{\Im z}$ bounded and $\abs{\Re z}\to\infty$. By Cauchy and taking limit we deform contour to the real axis

$\square$

Proof. Let $\curl{f_j}$ be a sequence of Schwartz functions that converges to $f$ in $L^{1}$. This convergence will then be uniform since

and uniform limit of continuous functions is continuous. Also observe $\norm{\ha{f}}{\infty}\leq\norm{f}{1}$$\square$

Proof. Given $\e>0,x$ fixed define

then by Theorem 12.4.5

Let $g(u)\define e^{-\pi\abs{u}^2}$ then

By Proposition 12.4.4(4)

$\square$

Proof. Observe by change of variables $u=x/\e$

then

By Minkowski’s inequality

where $f_{x}(y)\define f(x+y)$ and by Dominated Convergence Theorem we are done.$\square$

Proof. By Lemma 12.4.9$\square$

Proof. $L$ commute with translations then $c_{\al}$ are constants since $\pd^{\al}$ are invariant under translations. $(\ha{Lu})(\xi)=P(\xi)\ha{u}$ where $P(\xi)=\sums{}{\abs{\al}\leq k }c_{\al}(2\pi i\xi)^{\al}$. Since Fourier Transform commute with rotations, $P(\xi)$ must be radial in $\xi$ so $P$ is a function of $\abs{\xi}$. So $P(\xi)=\sums{}{}c_{,}m(2\pi i\abs{\xi})^{m}=\sums{}{j}c_{j}(2\pi i\abs{\xi})^{2j}=\sums{}{j}c_{j}(2\pi i\abs{\xi}^{2})^{j}=\sums{}{}a_{j}(\la)^{j}$$\square$

Proof. This is a simple corollary of Gauss Divergence Theorem in $\R^{n}$ ^[9]

Take $\vec{F}=v\vec{\nabla}u$ then $\vec{\nabla}(v\vec{\nabla}u)=\vec{\nabla}v\cd \vec{\nabla}u-v\la u$$\square$

Proof. Apply chain rule.$\square$

Proof. By Proposition 12.4.14 we have

the logarithmic derivative. Thus

So $\p'=const\cd r^{1-n}$ then integration finishes the proof.$\square$

Proof. It is clear that $N(x)$ is harmonic in $\R^{n}\del \curl{0}$ as it should be for $\la(T_{N})=\de$. We use “regularization” to make $N$ a $C^{\infty}$ function by “shifting” the “pole” at $x=0$ away into $\com^{n}$. Define

Write $N^{\e}$ for $T_{N^\e}$ then $N^{\e}\to N$ in $\cali{D}'$. First $N^{\e}\to N$ pointwise and $N^{\e},N$ are dominated for $\e\leq 1$ by $\abs{N}\in C^{\infty}_{c}\suf L^{1}_{loc}$. By Dominated Convergence theorem

for $\p\in\cali{D}=C_{c}^{\infty}$.

On radial functions

hence

where $\psi(x)=\frac{n}{\om_{n}}\parens{\abs{x}^2+1}^{-\frac{n+2}{2}}$. Observe that

so

By Lemma 12.4.10 with $g=\psi,f=\p$ then $\inner{\la N^\e}{\p}\to a\p(0)$ as $\e\to 0$. It remains to show that $a=1$.$\square$