= Solution
Consider the operators on all of $C_0(X)$ or $L^2(\mu)$, extending the displayed positive-function formula linearly. Here $C_0(X)$ is the <space of continuous functions vanishing at infinity>. The multiplier has modulus at most one, so $\|P_tf\|\leq\|f\|$, and multiplication of exponentials gives $P_{s+t}=P_sP_t$ and $P_0=I$. For <strong continuity> on $C_0(X)$, choose a compact $K$ outside which $|f|$ is small. On $K$ the continuous $k$ is bounded, making $e^{-tk}\to1$ uniform; outside $K$, $|(e^{-tk}-1)f|\leq|f|$. For $L^2(\mu)$, use <pointwise convergence> and the <dominated convergence theorem> with bound $|(e^{-tk}-1)f|^2\leq|f|^2$. Thus both are <contraction semigroups>, even when $k$ is unbounded.
The <generator of a multiplication semigroup> is
$$
\boxed{Lf=-kf,\qquad D_{C_0}(L)=\{f\in C_0(X):kf\in C_0(X)\},\qquad D_2(L)=\{f\in L^2(\mu):kf\in L^2(\mu)\}.}
$$
Indeed pointwise difference quotients converge to $-kf$. A <norm> limit therefore forces the displayed domain condition, using an almost-everywhere convergent subsequence for the $L^2$ case. Conversely $|(e^{-tk}-1)/t|\leq k$, so the <dominated convergence theorem> proves the $L^2$ limit when $kf\in L^2$. In $C_0(X)$ write the error as $kf$ times $1-(1-e^{-tk})/(tk)$, with the value at $k=0$ supplied by continuity. That second factor tends uniformly to zero where $k$ is bounded and has modulus at most one; the compact-set and small-tail argument applied to $kf$ proves <uniform convergence>. To prove directly that $L$ is a <closed operator>, take $f_n\to f$ and $Lf_n\to g$. Pointwise limits in $C_0$, or a common almost-everywhere convergent subsequence in $L^2$, give $g=-kf$. Thus $f$ is in the appropriate <generator domain> and $Lf=g$. In $L^2$ the domain is dense because $f\mathbf1_{\{k\leq n\}}\to f$; in $C_0$ compactly supported <continuous functions> are a dense subspace contained in the domain.
A real <multiplication operator> on its maximal $L^2$ domain is an <unbounded self-adjoint operator>. One direct verification is to test the adjoint relation against functions supported on $\{k\leq n\}$. If $h$ is in the <adjoint operator> domain with representing vector $g$, those tests force $g=-kh$ on each such set, hence $kh\in L^2$. The converse follows by integration. On the <sigma-finite measure> spaces used below, the <spectrum of a real multiplication operator> is its <essential range>; spectral values are detected by <unit vectors> supported where the multiplier is arbitrarily close to that value, while outside the essential range its reciprocal is a bounded resolvent multiplier.
For the <heat semigroup> on $L^2(\mathbb R,dx)$, take the normalization $L_H=f''$. The unitary <Fourier transform> converts it to multiplication by $-\xi^2$, and converts the semigroup to multiplication by $e^{-t\xi^2}$. Consequently
$$
\boxed{D(L_H)=H^2(\mathbb R),\qquad\sigma(L_H)=(-\infty,0],\qquad L_H=L_H^*.}
$$
The <Sobolev space> domain is exactly $\{f:\xi^2\widehat f\in L^2\}$. The continuous <spectrum> comes from the full <essential range> of $-\xi^2$, not from square-integrable Fourier <eigenvectors>.
For the <Ornstein-Uhlenbeck semigroup> on standard <Gaussian measure>, the normalized <Probabilists' Hermite polynomials> $h_n=\operatorname{He}_n/\sqrt{n!}$ form an <orthonormal basis>. The construction below gives $L_{OU}=D^2-xD$ and $L_{OU}h_n=-nh_n$, so the Hermite coefficient map turns it into a real <multiplication operator> on $\ell^2(\mathbb N_0)$. Therefore
$$
\boxed{D(L_{OU})=\left\{\sum_{n\geq0}c_nh_n:\sum_{n\geq0}n^2|c_n|^2<\infty\right\},\quad\sigma(L_{OU})=\{0,-1,-2,\ldots\},\quad L_{OU}=L_{OU}^*.}
$$
Finally, \b[the heat semigroup on the whole real line has no positive <spectral gap>], whereas \b[the standard <Ornstein-Uhlenbeck semigroup> has gap one]. For the first claim, take a nonzero smooth compactly supported $\phi$ and set $f_R(x)=R^{-1/2}\phi(x/R)$. Then $\|f_R\|_2^2=\|\phi\|_2^2$, but $\|f_R'\|_2^2=R^{-2}\|\phi'\|_2^2$. This disproves a uniform whole-line <Poincare inequality> $\|f\|_2^2\leq C\|f'\|_2^2$; one can also choose $\int\phi=0$. Lebesgue measure here is infinite, so it has no normalized probability mean to subtract. For <Gaussian measure>, the Hermite expansion instead gives the sharp <Gaussian Poincaré inequality>
$$
\boxed{\operatorname{Var}_\gamma(f)=\sum_{n\geq1}|c_n|^2\leq\sum_{n\geq1}n|c_n|^2=\int|f'|^2\,d\gamma.}
$$
Equality holds for affine functions. The identity on the right is interpreted on the energy-form domain, which is larger than the full <operator domain>.
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