Let ${\cal S}(\R^d)$ denote the Schwartz space on $\R^d$ and let $H_0f\colon=\sum\pa_j^4f$. Then $-H_0$ is a dissipative operator and $\im(1+H_0)$ is dense in $L_2(\R^d)$. Thus its closure $-H$ is the generator of a continuous contraction semigroup on $L_2(\R^d)$.
Since for $f,g\in{\cal S}(\R^d)$: $\int\pa_jf.g=-\int f.\pa_jg$ we get:
$$
\int H_0f.g\,d\l=\sum_j\int\pa_j^2f\pa_j^2g\,d\l,
$$
proving that $H_0$ is symmetric and positive. Thus its closure $H$ is symmetric and positive and for all $a > 0$:
$$
\int(af-H_0f).f\,d\l\geq a\norm f_2^2
\quad\mbox{i.e.}\quad
\norm{af-H_0f}_2\geq a\norm f_2
$$
which proves that $H_0$ is dissipative. Finally the Fourier transform of $g\colon=(1+H_0)f$ is $\wh g(y)=(1+\sum y_j^4)\wh f(y)$. Thus for any $g\in{\cal S}(\R^d)$ the function $f$ defined by
$$
\wh f(y)=(1+\sum y_j^4)^{-1}\wh g(y)
$$
is again in ${\cal S}(\R^d)$ and $(1+H_0)f=g$. Hence $\im(1+H_0)\spe{\cal S}(\R^d)$.