Dependencies

\(\mathcal{D}(\Omega ) = C_c^\infty (\Omega )\) is the set of test functions together with a topology determined by its converging sequences : \(\phi _n \to \phi \) in \(\mathcal{D}(\Omega )\) if

• there exists a compact subset \(K \subset \Omega \) such that \(\operatorname{Supp}(\phi _n) \subset K\).

• For all multiindices \(\alpha \) we have \(\partial _\alpha \phi _n \to \partial _\alpha \phi \) in uniformly.

Convention: if \(x \in \mathbb {R}^d \setminus \Omega \), then \(\phi (x) := 0\).
\(\mathcal{D}'(\Omega )\) is the topological dual space, i.e. the space of continuous linear functionals \(D(\Omega ) \to \mathbb {C}\) with the weak-*-convergence, i.e. pointwise convergence.

Let \(F \in \mathcal{D}'(\Omega ) , \psi \in D\). The following two distributions coincide:

1. The distribution determined by the smooth function \(x \mapsto F(\tau _x (\psi ^R))\).

2. The distribution \(\phi \mapsto F(\psi ^R * \phi )\).

We write this distribution as \(F * \psi \).

Let \(f,g \in L^1_{loc}(\Omega )\) such that \(\Lambda _f = \Lambda _g\). Then \(f = g\) almost everywhere.

Then \(C^\infty (\mathbb {R}^d)\) is dense in \(\mathcal{D}'(\mathbb {R}^d)\).

For a multiindex \(\alpha \) and a distribution \(F\) define the distribution

For \(F \in D', \phi \in D \) , We have

Consider the increasing sequence of norms on \(C^\infty (\Omega )\) defined by

with the obvious notion of convergence.

We have a continuous inclusions \(\mathcal{D}\subset \mathcal{S}\), hence \(\mathcal{S}' \subset \mathcal{D}'(\mathbb {R}^d)\).
Moreover, this inclusion is dense. Hence being tempered is a property of distributions.

If \(F \in \mathcal{D}'\) has compact support then it is tempered: Choose \(\eta \in \mathcal{D}\) such that \(\eta |_U = 1\) on some neighborhood \(U \supset \operatorname{Supp}(F)\). Then \(F(\eta \phi ) = F(\phi )\) for all \(\phi \in \mathcal{D}\) and \(\phi \mapsto F (\eta \phi )\) defines a continuous functional on \(\mathcal{S}\). It does not depend on the choice of \(\eta \), because given another such \(\eta '\) and any \(\phi \in \mathcal{S}\) we have \(\operatorname{Supp}((\eta - \eta ') \cdot \phi ) \cap \operatorname{Supp}(F) = \varnothing \).

• All \(\partial ^\alpha F\) are tempered.

• Let \(\psi \in C^\infty \) be slowly increasing, i.e. for each \(\alpha \) exists \(N_\alpha \) such that \(\partial _x^\alpha \psi (x) = O(|x|^{N_\alpha })\). Then \(\psi F\), defined by \((\psi F)(\phi )(F(\psi \phi )\) is tempered.

The fourier transformation is a continuous bijection

We have

Define

and similarly for the inverse transform \(f \mapsto \check{f}\).

A fundamental solution of \(L\) is a distribution \(F\) such that \(L(F) = \delta \)

The operator

defines an inverse to \(L\)

The characteristic polynomial of \(L\) is

For \(\lambda {\gt} -d\), Let \(H_\lambda \) be the tempered distribution associated to \(|x|^\lambda \in L^1_{loc}\).
If \(-d {\lt} \lambda {\lt} 0\) then

with

If \(d = 2\), the function \(F := 1/(2\pi ) \log |x| \in L^1_{loc}\) is a fundamental solution of \(\Delta \)

\(F\) is a fundamental solution of \(L = \frac{\partial }{\partial t} - \Delta _x\).

Let \(f\in L^2(V,E)\) and take a sequence \((f_n)_n\subset L^1(V,E)\cap L^2(V,E)\) with \(f_n\xrightarrow [L^2]{}f\). Set

the limit taken in the \(L^2\)-sense.

Let \(f\in L^1(V,E)\). Its Fourier transform (w.r.t. \(L\)) is the function \(\mathcal Ff=\widehat f:W\to E\) given by

The inverse Fourier transform (w.r.t. \(L\)) is similarly defined as

Define analogously \(\mathcal F^{-1}f:=\check f:=\lim _{n\to \infty }\check f_n\), if \(f_n\xrightarrow [L^2]{}f\in L^2(V,E)\) with \((f_n)_n\subset L^1(V,E)\cap L^2(V,E)\). By the same arguments as above, this is well-defined.

Let \(f\in L^1(V,E)\). Then its Fourier transform \(\widehat f\) is well-defined and bounded. In particular, the Fourier transform defines a map \(\mathcal F:L^1(V,E)\to L^\infty (V,E)\).

Let \(f\in L^1(V,E)\). Then \(\widehat f\) is continuous.

Let \(x\in V\) and \(\delta {\gt}0\). Define the modulated Gaussian

Its Fourier transform (w.r.t. the inner product) is given by

Let \(f,g\in L^1(V,E)\). Then

Lef \(f,g\in L^1(V,E)\), \(t\in \mathbb {R}\) and \(a,b\in \mathbb {C}\). The Fourier transform satisfies the following elementary properties:

1. \(\mathcal F(af+bg)=a\mathcal Ff+b\mathcal Fg\)(Linearity)

2. \(\mathcal F(f(x-t)) = e^{-2\pi i ty}\mathcal F f(y)\)(Shifting)

3. \(\mathcal F(f(tx)) = \frac1{|t|}\mathcal F f(\frac yt)\)(Scaling)

4. If \(E\) admits a conjugation, then \(\mathcal F(\overline{f(x)}) = \overline{\mathcal Ff(-y)}\)(Conjugation)

5. Define the convolution of \(f\) and \(g\) w.r.t. a bilinear map \(M:E\times E\to F\) as

   \[ (f\ast _Mg)(w):=\int _VM(f(v),g(w-v))\, d\mu (v). \]

   Then \(\mathcal F(f\ast _M g) =M(\mathcal Ff,\mathcal Fg)\) (Convolution)

Let \(f\in L^2(V,E)\) and \((f_n)_n\subset L^1(V,E)\cap L^2(V,E)\) a sequence with \(f_n\xrightarrow [L^2]{}f\). Then \((\widehat f_n)_n\) is a Cauchy sequence, hence converges in \(L^2(V,E)\).

This is well-defined: By Lemma 2.15, the limit exists. Further it does not depend on the choice of sequence \((f_n)_n\). If \(f\in L^1(V,E)\cap L^2(V,E)\), this definition agrees with the Fourier transform on \(L^1(V,E)\).

Let \(X=[0,2\pi ]\) with normalized Lebesgue measure \(\frac{d\theta }{2\pi }\) and let \(Y=\mathbb {Z}\) with counting measure.
Consider the operator \(T: f \mapsto \{ a_n\} _{n\in \mathbb {Z}}\) where

For \(1\leq p\leq 2\) and \(\frac{1}{p}+\frac{1}{q}=1\), we have

For \(1\leq p \leq 2\) and \(q\) conjugate exponent to \(p\), we have

Let \((X, \mu )\) be a measure space and \(0{\lt} p_0 {\lt}p_1\leq \infty \), \(0 \le p \le \infty \). Assume we have \(t \ge 0\) such that

Let \(f : X \to \mathbb C\) be a measurable function. Then

In particular, the \(f\) in \(L^{p_0}(X) \cap L^{p_1}(X)\), then \(f\) is in \(L^p(X)\).

\(L^1(V,E)\cap L^2(V,E)\) is dense in \(L^2(V,E)\).

Let \(p\) and \(q\) be real conjugate exponents. Let \(f\) be measurable. Then

In particular, if the right hand side formula is finite, \(f\in L^q\).

Let \(f\) be measurable and the measure \(\mu \) be \(\sigma \)-finite. Then

Let \(S\) be the strip \(S:=\{ z \in \mathbb {C}\ | \ 0 {\lt} \mathrm{Re} \, z {\lt} 1 \} \). Let \(f: \overline{S} \to \mathbb {C}\) be a function that is holomorphic on \(S\) and continuous and bounded on \(\overline{S}\).
Assume \(M_0, M_1\) are positive real numbers such that for all values of \(y\) in \(\mathbb {R}\), we have

i.e., the absolute values of the function on the lines \(\{ \mathrm{Re} \, z = 0\} \) and \(\{ \mathrm{Re} \, z = 1\} \) are bounded by \(M_0\) and \(M_1\) respectively.
Then, for all \(0 \leq t \leq 1\) and for all real values of \(y\), we have

Let \(K_\delta (v)=\delta ^{-n/2}e^{-\pi |v|^2/\delta }\) as in Lemma 2.6. This is a good kernel, called the Weierstrass kernel, satisfying

Furthermore, it satisfies the stronger bounds

for some constant \(B\) independent of \(\delta \).

Convolution \(F * \psi \) is continuous in both variables.

Let \(\phi \in C_c^{\infty }(\Omega \times \Omega )\). Then for any \(F \in D'(\Omega )\) we have

Let \(f \in L^1_{loc}(\mathbb {R}^d)\) such that there exists \(N \ge 0\) with

The distribution \(\Lambda _f\) is tempered.

There exists a sequence \(\psi _n \in C_c^{\infty }(\Omega )\) such that \(\Lambda \psi _n \to \delta _0\) in \(D'(\Omega )\).

Let \(f:V\to E\) be integrable and continuous. Assume \(\widehat f\) is integrable as well. Then

Let \(f\in L^1(V,E)\). If \(\widehat f\in L^1(V,E)\), then \(\mathcal F^{-1}\mathcal Ff=f\).

The Fourier transform induces a continuous linear map \(L^2(V,E) \to L^2(V,E)\).

Plancherel’s Theorem, the inversion formula, and the properties of Lemma 2.5 hold for the Fourier transform on \(L^2(V,E)\) as well.

Let \(f:V\to E\) be integrable. Let \(K_\delta \) be the Weierstrass kernel from Lemma 2.7, or indeed any family of functions satisfying the conditions of Lemma 2.7. Then

in the \(L^1\)-norm. If \(f\) is continuous, the convergence also holds pointwise.

Let \(U\) be a connected open set in a complex normed space \(E\). Let \(f:E\to F\) be a function that complex differentiable on \(U\) and continuous on \(\bar{U}\).
If \(|f(z)|\) takes its maximum on a point \(u\in U\), then it must be constant on \(\bar{U}\).

Suppose that \(f : V \to E\) is in \(L^1(V,E)\cap L^2(V,E)\) and let \(\widehat{f}\) be the Fourier transform of \(f\). Then \(\widehat{f},\check{f}\in L^2(V,E)\) and

Suppose that \(f : V \to E\) is in \(L^1(V,E)\cap L^2(V,E)\) and let \(\widehat{f}\) be the Fourier transform of \(f\). Then \(\widehat{f},\check{f}\in L^2(V,E)\) and

Let \((X, \mu )\) and \((Y, \nu )\) be measure spaces and consider all \(L^p\) spaces to be complex valued.
Suppose \(T\) is a linear map \(L^{p_0}(X) + L^{p_1}(X) \to L^{q_0}(Y) + L^{q_1}(Y)\) that restricts to bounded operators \(L^{p_0} \to L^{q_0}\) and \(L^{p_1} \to L^{q_1}\). Let \(M_0, M_1\) be the respective bounds, i.e.,

Then, for any pair \((p, q)\) such that there is a \(t\) in \([0,1]\) for which

we have that the operator is bounded \(L^p \to L^q\), and in particular