5. The Fourier–Laplace Transformation and Material Law Operators

For \(f\in L_{1}(\mathbb {R};H)\) we define the Fourier transform \(\widehat {f}\) of f by

Let \(f\in L_{1}(\mathbb {R};H)\) . Then \(\widehat {f}\in C_{\mathrm {b}}(\mathbb {R};H)\) and \(\lim _{\left \vert t \right \vert \to \infty } \big \|\widehat {f}(t)\big \| = 0\) . Moreover,

First, note that \(\widehat {f}\) is continuous by dominated convergence and bounded with This shows that the mapping defines a bounded linear operator. Moreover, for \(\varphi \in C_{\mathrm {c}}^1(\mathbb {R};H)\) we compute for s≠0 and thus, which shows that \(\lim _{\left \vert s \right \vert \to \infty }\left \Vert \widehat {\varphi }(s) \right \Vert =0\). By the facts that \(C_{\mathrm {c}}^1(\mathbb {R};H)\) is dense in \(L_{1}(\mathbb {R};H)\) (see Lemma 3.​1.​8), \(\left \{ f\in C_{\mathrm {b}}(\mathbb {R};H) \,;\, \lim _{\left \vert t \right \vert \to \infty }\left \Vert f(t) \right \Vert =0 \right \}\) is a closed subspace of \(C_{\mathrm {b}}(\mathbb {R};H)\) and the operator in (5.1) is bounded, the assertion follows. □

We define to be the Schwartz space of rapidly decreasing functions on \(\mathbb {R}\) with values in H.

\(\mathcal {S}(\mathbb {R};H)\) is a Fréchet space with respect to the seminorms Moreover, \(\mathcal {S}(\mathbb {R};H)\subseteq \bigcap _{p\in \left [1,\infty \right ]}L_{p}(\mathbb {R};H)\). Indeed, \(\mathcal {S}(\mathbb {R};H)\subseteq L_{\infty }(\mathbb {R};H)\) by definition, and for \( f \in \mathcal {S}(\mathbb {R};H)\) and \(1\leqslant p<\infty \) we have that

For \(f\in \mathcal {S}(\mathbb {R};H)\) we have \(\widehat {f}\in \mathcal {S}(\mathbb {R};H)\) and the mapping

is bijective. Moreover, for \(f,g\in L_{1}(\mathbb {R};H)\) we have that

Additionally, if \(f,\widehat {f}\in L_{1}(\mathbb {R};H)\) then

Let \(f\in \mathcal {S}(\mathbb {R};H)\). By Exercise 5.1 we have and Using these formulas, one can show that \(\widehat {f}\in \mathcal {S}(\mathbb {R};H).\) Since the bijectivity of the Fourier transformation on \(\mathcal {S}(\mathbb {R};H)\) would follow from (5.3), it suffices to prove the formulas (5.2) and (5.3). Let \(f,g\in L_{1}(\mathbb {R};H).\) Then we compute using Proposition 3.​1.​6 and Fubini’s theorem which yields (5.2). For proving formula (5.3), we consider the function γ defined by for \(t\in \mathbb {R}\). Clearly, \(\gamma \in \mathcal {S}(\mathbb {R})\). We claim that \(\widehat {\gamma }=\gamma \). Indeed, we observe that γ solves the initial value problem y′ + ty = 0 subject to y(0) = 1; if we can show that \(\widehat {\gamma }\) solves the same initial value problem, then their equality would follow from the uniqueness of the solution. First, we observe that \(\widehat {\gamma }(0)=\frac {1}{\sqrt {2\pi }}\int _{\mathbb {R}}\mathrm {e}^{-\frac {t^{2}}{2}}\,\mathrm {d} t=1.\) Second, we compute using the formulas (5.4) and (5.5) that Altogether, we have shown that \(\widehat {\gamma }\) solves the same initial value problem as γ and hence, \(\widehat {\gamma }=\gamma \). Let now \(f\in L_{1}(\mathbb {R};H)\) with \(\widehat {f}\in L_{1}(\mathbb {R};H),\) a > 0 and x ∈ H. Then we compute using (5.2) for each \(s\in \mathbb {R}\). Since this holds for all x ∈ H we get Letting a → 0 in the latter equality, we obtain where we have used dominated convergence for the term on the left-hand side. In order to compute the limit on the right-hand side, we first observe that and hence, for each a > 0 the operator is bounded by \(\left \Vert \gamma \right \Vert { }_{1}\). Moreover, since \(S_{a}\psi \to \psi (\cdot )\left \Vert \gamma \right \Vert { }_{1}\) as a → 0 for \(\psi \in C_{\mathrm {c}}(\mathbb {R};H)\), we infer that for each \(f\in L_{1}(\mathbb {R};H)\). Hence, passing to a suitable sequence \(\left (a_{n}\right )_{n}\) in \(\mathbb {R}_{>0}\) tending to 0, we get Using this identity for the right-hand side of (5.6), we get and since \(\left \Vert \gamma \right \Vert { }_{1}=\sqrt {2\pi }\), we derive (5.3). □

The mapping

extends to a unitary operator on \(L_2(\mathbb {R};H)\) , again denoted by \(\mathcal {F}\) , the Fourier transformation . Moreover, \(\mathcal {F}^{\ast }=\mathcal {F}^{-1}\) is given by \(f\mapsto \widehat {f}(-\cdot )\).

Using (5.2) and (5.3) we obtain that for all \(f,g\in \mathcal {S}(\mathbb {R};H)\) and thus, in particular, Moreover, \(\operatorname {dom}(\mathcal {F})=\operatorname {ran}(\mathcal {F})=\mathcal {S}(\mathbb {R};H)\) is dense in \(L_2(\mathbb {R};H)\) and hence, the first assertion follows by Exercise 5.2. As \(\mathcal {F}\) is unitary, we have \(\mathcal {F}^*=\mathcal {F}^{-1}\), thus, by (5.2) applied to \(f,g\in \mathcal {S}(\mathbb {R};H)\), we read off (using Proposition 2.​3.​8) that \(\mathcal {F}^{-1}=(f\mapsto \widehat {f}(-\cdot ))\), which yields all the claims of the theorem at hand. □

We emphasise that for \(f\in L_2(\mathbb {R};H)\) the Fourier transform \(\mathcal {F}f\) is not given by the integral expression for L ₁-functions, simply because the integral does not need to exist. However, by dominated convergence where the limit is taken in \(L_2(\mathbb {R};H).\)

Let \(\nu \in \mathbb {R}\). We define the Fourier–Laplace transformation as

Note that \(\mathcal {L}_{\nu }=\mathcal {F}\exp (-\nu \mathrm {m})\) is unitary as an operator from \(L_{2,\nu }(\mathbb {R};H)\) to \(L_2(\mathbb {R};H)\) since it is the composition of two unitary operators. For \(\varphi \in C_{\mathrm {c}}^\infty (\mathbb {R};H)\), we have the expression which shows that \(\mathcal {L}_{\nu }\) can be interpreted as a shifted variant of the Fourier transformation, where the real part in the exponent equals ν instead of zero.

Let \(V\colon \mathbb {R}\to \mathbb {K}\) be measurable. We define the multiplication-by-V operator as with In particular, if V  is the identity on \(\mathbb {R}\) we will just write m instead of \(\operatorname {id}(\mathrm {m})\) and call it the multiplication-by-the-argument operator.

Note that the multiplication-by-V  operator is a vector-valued analogue of the multiplication operator seen in Theorems 2.​4.​3 and 2.​4.​7. The statements in these theorems generalise (easily) to the vector-valued situation at hand. Thus, as in Theorem 2.​4.​3, one shows that m is selfadjoint. Moreover, when H ≠ {0}, in a similar fashion to the arguments carried out in Theorem 2.​4.​7 one shows that

Let \(\nu \in \mathbb {R}\) . Then

In particular,

We first prove the assertion for ν≠0 and show that The assertion will then follow by Theorem 2.​4.​3(d). Note that \(\frac {1}{\mathrm {i}\mathrm {m}+\nu }\in L(L_2(\mathbb {R};H))\) by Proposition 2.​4.​6, and hence, both operators I _ν and \(\mathcal {L}_{\nu }^{\ast }(\tfrac {1}{\mathrm {i}\mathrm {m}+\nu })\mathcal {L}_{\nu }\) are bounded and defined on the whole of \(L_{2,\nu }(\mathbb {R};H).\) Thus, it suffices to prove the equality on a dense subset of \(L_{2,\nu }(\mathbb {R};H)\), like \(C_{\mathrm {c}}(\mathbb {R};H).\) We will just do the computation for the case when ν > 0. So, let \(\varphi \in C_{\mathrm {c}}(\mathbb {R};H)\) and compute for \(t\in \mathbb {R}.\) For ν < 0 the computation is analogous. In the case when ν = 0 we observe that

□

A mapping \(M\colon \operatorname {dom}(M)\subseteq \mathbb {C}\to L(H)\) is called a material law if Moreover, we set to be the abscissa of boundedness of M.

Let us state various examples of material laws.

Let \(M\colon \operatorname {dom}(M)\subseteq \mathbb {C}\to L(H)\) be a material law. Then, for \(\nu >\mathrm {s}_{\mathrm {b}}\left ( M \right )\) , the operator

is bounded. Moreover, we define the material law operator

and obtain

The proof is clear. □

The set of material laws is an algebra and the mapping of assigning a material law to its corresponding material law operator is an algebra homomorphism in the following sense. For j ∈{1, 2} let \(M_j\colon \operatorname {dom}(M_j)\subseteq \mathbb {C} \to L(H)\) be material laws, \(\lambda \in \mathbb {C}\). Then M ₁ + M ₂ (with domain \(\operatorname {dom}(M_1)\cap \operatorname {dom}(M_2)\)), λM ₁ and M ₁ ⋅ M ₂ (with domain \(\operatorname {dom}(M_1)\cap \operatorname {dom}(M_2)\)) are material laws as well. Moreover, \(\mathrm {s}_{\mathrm {b}}\left ( M_1+M_2 \right ), \mathrm {s}_{\mathrm {b}}\left ( M_1\cdot M_2 \right )\leqslant \max \{\mathrm {s}_{\mathrm {b}}\left ( M_1 \right ),\mathrm {s}_{\mathrm {b}}\left ( M_2 \right )\}\). Furthermore, if M ₂(z) is a scalar for all \(z\in \operatorname {dom}(M_2)\), then for \(\nu >\max \{\mathrm {s}_{\mathrm {b}}\left ( M_1 \right ),\mathrm {s}_{\mathrm {b}}\left ( M_2 \right )\}\) we have (M ₁ M ₂)(∂ _t,ν) = M ₁(∂ _t,ν)M ₂(∂ _t,ν) = M ₂(∂ _t,ν)M ₁(∂ _t,ν) = (M ₂ M ₁)(∂ _t,ν).

We now revisit the material laws presented in Example 5.3.1 and compute their corresponding operators, M(∂ _t,ν).

Let \(\mu ,\nu \in \mathbb {R}\) with μ < ν, and set . Let \(g\colon \overline {U}\to H\) be continuous and holomorphic on U such that \(g(\mathrm {i}\cdot +\nu ),g(\mathrm {i}\cdot +\mu )\in L_2(\mathbb {R};H)\) and there exists a sequence \((R_{n})_{n\in \mathbb {N}}\) in \(\mathbb {R}_{\geqslant 0}\) such that R _n →∞ and

Then

Let \(t\in \mathbb {R}\). By Cauchy’s integral theorem, we have that where \(\gamma _{R_{n}}\) is the rectangular closed path with corners ±iR _n + μ, ±iR _n + ν (see Fig. 5.1). Thus, we have that Note that with the help of the formula for the inverse Fourier transformation (see Theorem 5.1.4) and \(\mathcal {L}_\nu ^*=(\mathcal {F}\exp (-\nu \mathrm {m}))^*=\exp (-\nu \mathrm {m} )^{-1}\mathcal {F}^*\) the left-hand side of (5.10) is nothing but and hence, there is a subsequence of (R _n)_n (which we do not relabel) such that the left-hand side of (5.10) tends to for almost every \(t\in \mathbb {R}\) as n →∞. As such, all we need to show is that the right-hand side of (5.10) tends to 0 as n →∞, which obviously follows by (5.9). □

Let \(M\colon \operatorname {dom}(M)\subseteq \mathbb {C}\to L(H)\) be a material law. Then, for \(\mu ,\nu >\mathrm {s}_{\mathrm {b}}\left ( M \right )\) and \(f\in L_{2,\nu }(\mathbb {R};H)\cap L_{2,\mu }(\mathbb {R};H)\) , we have

Moreover, M(∂ _t,ν) is causal for all \(\nu >\mathrm {s}_{\mathrm {b}}\left ( M \right )\).

Let μ < ν. We prove the assertion for with a < b and x ∈ H first. For \(\rho \in \mathbb {R}\) we compute for all \(t\in \mathbb {R}\setminus \{0\}\). Moreover, we define and prove that g satisfies the assumptions of Lemma 5.3.5. First, we note that g is bounded on \(\left \{ z\in \mathbb {C} \,;\, \mu \leqslant \operatorname {Re} z\leqslant \nu \right \}\setminus \{0\}\). Indeed, we only need to prove that it is bounded near 0 provided that μ⩽0. To that end, we observe Thus, g is bounded near 0. In particular, z = 0 is a removable singularity and, hence, g can be extended holomorphically to \(\mathbb {C}_{\operatorname {Re}\geqslant \mu }\). Moreover, for \(\rho \geqslant \mu \) we have that The first term on the right-hand side is finite since g is bounded, while the second term can be estimated by This proves that \(g(\mathrm {i}\cdot +\rho )\in L_2(\mathbb {R};H)\) for each \(\rho \geqslant \mu \) and hence, particularly for ρ = μ and ρ = ν. Finally, for \(\rho \geqslant \mu \) we have that which together with the boundedness of g yields (5.9) by dominated convergence. This shows that g satisfies the assumptions of Lemma 5.3.5 and thus By linearity, this equality extends to \(S_{\mathrm {c}}(\mathbb {R};H)\) and so, is well-defined. Moreover, F is uniformly Lipschitz continuous (observe that \(\sup _{\nu \geqslant \mu } \left \Vert F^{\nu } \right \Vert \leq \left \Vert M \right \Vert { }_{\infty ,\mathbb {C}_{\operatorname {Re}>\mu }}\)) and hence, the assertions follow from Lemma 4.​2.​5. □

Let ( Ω,  Σ, μ) be a σ-finite measure space, X a Banach space and \(I\subseteq \mathbb {R}\) an open interval. Let g: I × Ω→ X such that g(t, ⋅) ∈ L ₁(μ;X) for each t ∈ I, and define

Let H ₀, H ₁ be two Hilbert spaces and \(U\colon \operatorname {dom}(U)\subseteq H_{0}\to H_{1}\) linear such that Show that U can be uniquely extended to a unitary operator between H ₀ and H ₁.

Let \(\Omega \subseteq \mathbb {C}\) be open, X a complex Banach space and f :  Ω → X. Prove that the following statements are equivalent: Assume now that X = L(X ₁, X ₂) for two complex Banach spaces X ₁, X ₂, let D ₁ ⊆ X ₁ be dense and \(D_{2}\subseteq X_{2}^{\prime }\) norming for X ₂. Prove that the statements (i) to (iv) are equivalent to Hint: For the difficult implications one might also consult [6, Appendix A]. In the same source one can find that in part (iii) it is enough for D to be separating.

Let \(\nu \in \mathbb {R}\) and \(k\in L_{1,\nu }(\mathbb {R})\). Prove that for \(f\in L_{2,\nu }(\mathbb {R};H)\).

Let α > 0 and define for \(t\in \mathbb {R}.\) Show that \(g_{\alpha }\in L_{1,\nu }(\mathbb {R})\) for each ν > 0 and that Use this formula and Exercise 5.4 to derive (5.8).

Hint: To compute the Fourier–Laplace transform of g _α, derive that \(\mathcal {L}_{\nu }g_{\alpha }\) solves a first order ordinary differential equation and use separation of variables to solve this equation.

Let \(\mu ,\nu \in \mathbb {R}\) with μ < ν and \(f\in L_{2,\nu }(\mathbb {R};H)\cap L_{2,\mu }(\mathbb {R};H)\). Moreover, set . Show that \(f\in \bigcap _{\mu <\rho <\nu }L_{2,\rho }(\mathbb {R};H)\cap L_{1,\rho }(\mathbb {R};H)\) and that is holomorphic.

Let H ₀, H ₁ be Hilbert spaces and \(T\colon L_{2,\nu }(\mathbb {R};H_{0})\to L_{2,\nu }(\mathbb {R};H_{1})\) linear and bounded. We call T autonomous if Tτ _h = τ _h T for each \(h\in \mathbb {R}\) (τ _h denotes the translation operator defined in Example 5.3.4). Prove that for autonomous T, the following statements are equivalent: Moreover, prove that for a material law M, the operator M(∂ _t,ν) is autonomous for each \(\nu >\mathrm {s}_{\mathrm {b}}\left ( M \right )\).