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For the physical-space L2 norm use $\Vert h{\Vert }_2^2={\mathbb{E}}_x|h(x){|}^2$. By part (i) and the Parseval identity on a finite group,Split this sum into the large spectrum $K$ and its complement. Because $u$ belongs to the Bohr set in the question, for $r\in K$,Consequently the contribution from $K$ is at most ${\epsilon }^2{\sum }_r|\hat{a}(r){|}^4\le {\epsilon }^2{\alpha }^3$ by part (ii). Outside $K$, $|\hat{a}(r)|<\theta$, and $|1-{\omega }^{-ru}|\le 2$. Thus that contribution is at mostAdding the estimates provesThe argument also covers empty $A$ or empty $K$. As usual the radius and threshold are nonnegative; a negative radius makes the premise empty whenever $K$ is nonempty. The estimate expresses Bohr-set almost periodicity of a convolution: a translation of a function by an element of the Bohr set barely changes the large Fourier coefficients on a finite abelian group, while the small ones have little total energy.

Use normalized Fourier analysis on a finite abelian group, identifying $A$ with its indicator function $a=1_A$ on the cyclic group ${\mathbb{Z}}_n$. The conventions areThe last operation is normalized convolution on a finite group. With these conventions the Fourier coefficients on a finite abelian group use a normalized average, while sums over frequencies use counting measure.

Expanding the normalized convolution on a finite group and putting $z=x-y$ givesFor the translation of a function $g(x)=f(x-u)$, putting $z=x-u$ yieldsThe two transforms are thereforeThe negative sign in the translation factor follows from the negative sign in our Fourier coefficient on a finite abelian group convention.

A contraction semigroup on a Banach space $X$ consists of bounded linear operators $S(t)$ with $S(0)=I$, $S(t+s)=S(t)S(s)$, $\Vert S(t)\Vert \le 1$, and $S(t)x\to x$ as $t\downarrow 0$ for each $x\in X$. Its infinitesimal generator of a semigroup isFor density, Yosida averaging of a semigroup gives $x_h=h^{-1}{\int }_0^{h}S(s)xds$. Taking the difference quotient of this Bochner integral shows that $x_h\in D(A)$ and $A{x}_h=(S(h)x-x)/h$. Strong continuity gives $x_h\to x$, so $D(A)$ is dense.

For closedness, the semigroup restricted to its generator domain satisfiesIf $x_n\to x$ and $A{x}_n\to y$, pass to the limit in this identity. After dividing by $t$, strong continuity gives $(S(t)x-x)/t\to y$. Hence $x\in D(A)$ and $Ax=y$. This proves generator of a strongly continuous semigroup is closed and densely defined.

The contraction form of the Hille-Yosida theorem states that a linear operator $A$ generates a contraction semigroup if and only if it is densely defined and closed, every real $\lambda >0$ lies in its resolvent set, andHere the $n=1$ bound implies the others by taking powers of the same bounded resolvent.

The H1 space is $H^1(\mathbb{R})=\{u\in L^2:u^{\prime }\in L^2\}$ with weak derivative $u^{\prime }$ and squared norm $\Vert u{\Vert }_2^2+\Vert u^{\prime }{\Vert }_2^2$. The displayed weighted space is the one-dimensional harmonic oscillator form domain, with inner productIf $u_n$ is Cauchy in $X$, it converges in $H^1$ to $u$, and $x{u}_n$ converges in L2 space to some $w$. On each bounded interval, multiplication by $x$ is bounded, so $w=xu$ there. Thus $xu\in L^2$ and convergence holds in $X$, proving completeness and the Hilbert space property. Equivalently this is closedness of the multiplication operator by $x$.

For the Sobolev characterization by bounded difference quotients, if $f\in H^1$ thenThe last assertion uses continuity of translation of a function in L2 space. Conversely, if $f\in L^2$ and the quotients are uniformly bounded for $0<|h|<1$, take a weakly convergent subsequence as $h\to 0$. For every test function $v$,The weak limit is therefore the weak derivative $f^{\prime }\in L^2$. This proves the characterization.

For $f\in H^1(\mathbb{R})$, its difference quotient equals $D_{h}f={\int }_0^1{f}^{\prime }(\cdot +\theta h)d\theta$ in L2 space. Thus $\Vert D_{h}f{\Vert }_2\le \Vert f^{\prime }{\Vert }_2$, and translation of a function gives $D_{h}f\to f^{\prime }$ in L2 space. Conversely, bounded difference quotients have a weakly convergent subsequence as $h\to 0$. Against a test function, $\int D_{h}fv\to -\int f{v}^{\prime }$, so that weak limit is the weak derivative of $f$. It lies in L2 space, which is precisely the first-order Sobolev space condition.