= Solution
A <contraction semigroup> on a <Banach space> $X$ consists of bounded linear operators $S(t)$ with $S(0)=I$, $S(t+s)=S(t)S(s)$, $\|S(t)\|\leq1$, and $S(t)x\to x$ as $t\downarrow0$ for each $x\in X$. Its <infinitesimal generator of a semigroup> is
$$
D(A)=\left\{x:\lim_{h\downarrow0}\frac{S(h)x-x}{h}\text{ exists in }X\right\},
\qquad
Ax=\lim_{h\downarrow0}\frac{S(h)x-x}{h}.
$$
For density, <Yosida averaging of a semigroup> gives $x_h=h^{-1}\int_0^hS(s)x\,ds$. Taking the difference quotient of this <Bochner integral> shows that $x_h\in D(A)$ and $Ax_h=(S(h)x-x)/h$. Strong continuity gives $x_h\to x$, so $D(A)$ is dense.
For closedness, the <semigroup restricted to its generator domain> satisfies
$$
S(t)x-x=\int_0^tS(s)Ax\,ds\qquad(x\in D(A)).
$$
If $x_n\to x$ and $Ax_n\to y$, 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>, and
$$
\|(\lambda-A)^{-n}\|\leq\lambda^{-n}\qquad(n\geq1).
$$
Here the $n=1$ bound implies the others by taking powers of the same bounded <resolvent>.
The <H1 space> is $H^1(\mathbb R)=\{u\in L^2:u'\in L^2\}$ with <weak derivative> $u'$ and squared norm $\|u\|_2^2+\|u'\|_2^2$. The displayed weighted space is the <one-dimensional harmonic oscillator form domain>, with inner product
$$
\langle u,v\rangle_X=\int_{\mathbb R}u\bar v+u'\overline{v'}+x^2u\bar v\,dx.
$$
If $u_n$ is Cauchy in $X$, it converges in $H^1$ to $u$, and $xu_n$ 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 $f\in H^1$ then
$$
D_hf=\int_0^1f'(\cdot+\theta h)\,d\theta,
\qquad
\|D_hf\|_2\leq\|f'\|_2,
\qquad
D_hf\longrightarrow f'\text{ in }L^2.
$$
The 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 $h\to0$. For every <test function> $v$,
$$
\int D_hf\,v\,dx=\int f(x)\frac{v(x-h)-v(x)}h\,dx\longrightarrow-\int f v'\,dx.
$$
The weak limit is therefore the <weak derivative> $f'\in L^2$. This proves the characterization.
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