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Under the Fourier transform, the operator $A(t)={\partial }_x^3+\varphi (t){\partial }_x$ is the Fourier multiplier operatorIts symbol is purely imaginary because $\varphi$ is real. Consequently $A(t)$ is skew-adjoint on $L^2(\mathbb{R})$ with common domain $H^3(\mathbb{R})$ and generates the strongly continuous unitary groupThe Plancherel theorem gives $\Vert e^{rA(t)}u{\Vert }_2=\Vert u{\Vert }_2$, so every $A(t)\in \mathcal{G}(1,0)$. Products of the frozen groups are also unitary; hence this is a stable family of semigroup generators with constants $1,0$.

For $u\in H^3(\mathbb{R})$,Since $\varphi \in C^1(\mathbb{R})$, the map $t\mapsto A(t)$ is continuously differentiable from $H^3$ to $L^2$. All hypotheses from part f are satisfied, so an evolution family exists on every finite interval $[0,T]$.

In this commuting Fourier multiplier operator example the solution operator can also be written explicitly:Its multiplier has absolute value one, directly confirming strong continuity, the evolution family law, preservation of $H^3$, and the required derivatives. The equation combines the dispersive Airy equation with a time-dependent linear transport equation.

Use the conventionThe Riemann-Lebesgue lemma states that if $f\in L^1({\mathbb{R}}^n)$, then $\hat{f}$ is continuous andContinuity follows from the dominated convergence theorem, since $e^{-ix\cdot {\xi }_j}\to e^{-ix\cdot \xi }$ and the integrands are dominated by $|f(x)|$.

For the decay, first take $g\in C_c^1({\mathbb{R}}^n)$. Choose a coordinate $j$ for which $|{\xi }_j|\ge |\xi |/\sqrt{n}$. Integration by parts givesand henceSince $C_c^1({\mathbb{R}}^n)$ is dense in $L^1({\mathbb{R}}^n)$, choose $g$ with $\Vert f-g{\Vert }_1<\epsilon$. The elementary Fourier bound givesso the decay for $g$ implies the decay for $f$.

The Parseval identity says that for $f,g\in L^2({\mathbb{R}}^n)$, with their Fourier transforms defined by the Plancherel theorem,In particular,

For the given radial function, as $|x|\to 0$,while as $|x|\to \infty$,Using polar coordinates, local integrability of $|f{|}^q$ is therefore determined byand integrability at infinity byThe assumptions $2a>-n$ and $b>n$ imply both $f\in L^2({\mathbb{R}}^n)$ and $f\in L^1({\mathbb{R}}^n)$. The Riemann--Lebesgue lemma and the direct estimate $|\hat{f}|\le \Vert f{\Vert }_1$ give $\hat{f}\in L^{\infty }$, while Parseval gives $\hat{f}\in L^2$. Finally, for every $2\le p<\infty$,Thus

For $s\in \mathbb{R}$, the Sobolev space $H^s({\mathbb{R}}^n)$ consists of tempered distributions $u$ such thatFor a multi-index $\alpha$,Sincewe obtain the Sobolev derivative estimateThus differentiation is bounded and linear between the stated spaces.

Taking the Fourier transform ofgivesThe only possible solution is thereforeMoreover,This proves existence, uniqueness, and bounded dependence. Conversely, $u\mapsto (-\Delta +1)u$ maps $H^{s+2}$ boundedly to $H^s$, sois a linear isomorphism with inverse multiplier $(1+|\xi {|}^2{)}^{-1}$. This is the massive Laplacian isomorphism on Sobolev spaces.

Finally, the assumed estimate and Plancherel theorem implyThus $(u_j)$ is bounded in $H^2({\mathbb{R}}^n)$. The functions and all their first derivatives are supported in the fixed bounded set $\overline{\Omega }$. Their $H^1$ bounds give uniform translation estimatesfor $v=u_j$ and $v=D_{\ell }{u}_j$. Tight support and the Rellich-Kondrashov compactness theorem for H01, equivalently the Fourier compactness criterion, therefore provide a common subsequence for which $u_j$ and every first derivative converge strongly in $L^2$. HenceThis is compactness from bounded support and an H2 bound.

Use the conventionThe Fourier inversion theorem says that, when $f,\hat{f}\in L^1({\mathbb{R}}^d)$,for almost every $x$.

Sincethe Fubini's theorem and Fourier inversion giveThus the Plancherel theorem identity in this convention is

The inverse integral is a continuous function of $x$, since $\hat{f}\in L^1$ permits dominated convergence. If $f$ is continuous, it and the inverse integral are continuous and agree almost everywhere. They must agree everywhere: otherwise their continuous difference would be nonzero on an open set of positive measure. This is the continuous version of Fourier inversion.

PutThis function is bounded near zero and is $O(u^{-2})$ at infinity, so $H\in L^1(\mathbb{R})$ and the defining integral for $g$ converges absolutely.

For the triangular functiondirect integration givesThe Fourier inversion theorem, or equivalently the Fourier transform of a triangular function, therefore yieldsIn particular $g(x)=0$ whenever $|x|>1$. Finally,This also agrees with the Plancherel theorem applied to $h$ and $H$.

Extend $u_k$ and $u$ by zero outside the bounded set $U$. Weak convergence makes $(u_k)$ bounded in $L^2(U)$. For each fixed $p$,because $e^{-ip\cdot x}\in L^2(U)$.

Fix $t>0$. On the finite-measure ball $|p|\le t$, the Fourier transforms are uniformly bounded byDominated convergence therefore givesOn the complementary region,whose integral tends to zero uniformly in $k$ as $t\to \infty$ by hypothesis. Splitting first at large $t$ and then taking $k\to \infty$, the Plancherel theorem yieldsThis is the same low-frequency compactness mechanism used in the Fourier proof of Rellich compactness.

For Schwartz $f,g$, Fourier inversion and Fubini giveTaking $g=f$ proves Plancherel theorem; polarization gives the full identity.

If $g\in L^2$ is orthogonal to every translate of $f$, then the correlation $g\ast \tilde{f}$ vanishes. By convolution theorem,
$\hat{g}\overline{\hat{f}}=0$ almost everywhere. Since $\hat{f}$ vanishes only at one point, a null set, $\hat{g}=0$ almost everywhere. By Plancherel theorem, $g=0$. Orthogonal-complement characterization proves density in $L^2$.

For a character of modulus $p^k$, a primitive Dirichlet character attains the bound by finite Plancherel theorem. An imprimitive character with $k\ge 2$ has periodic values modulo $p^{k-1}$, so its frequency-one sum vanishes. The principal Dirichlet character modulo $p$ has unnormalized sum $-1$. The sign convention of the exponential does not affect its magnitude.

With the unnormalized Fourier transform, the scaling function with transform $1_{[-\pi ,\pi )}$ is $\varphi (x)=\sin (\pi x)/(\pi x)$. Its integer translates are orthonormal by orthonormal translates and Fourier periodization. Its refinement mask is the periodic indicator of $[-\pi /2,\pi /2)$, with $a_0=1$ and $a_n=2\sin (n\pi /2)/(\pi n)$. It belongs to $L^2$ but not $L^1$, so its transform is naturally interpreted through the Plancherel theorem.

With the standard Fourier convention, Plancherel theorem applied to $1_{[-1,1]}$ gives