# Spectrum of convolution operator on bounded continuous functions (on R^d)

Hi!

Let $a(x)\geq0$, $x\in R^d$ and $\int_{R^d} a(x) dx=1$. Then the operator $Af = a*f -f$ is bounded on the space of continuous functions on $R^d$ vanishing at infinity and it is a generator of a Markov semigroup, therefore, its spectrum will be only on the left half-plane (and bounded, of course).

If we consider now the Banach space of all bounded continuous functions on $R^d$, then $A$ will be bounded as well, however, what can we say about left half plane. Could it be shown that this operator is dissipative?

If it's dissipative on $C_0({\bf R}^d)$ then it should be dissipative on $C_b({\bf R}^d)$ too.
First check this in the case that $a$ has bounded support. The point is that if $a$ is supported on a ball of radius $r$ about the origin then $Af(x)$ only depends on values of $f(y)$ for $|x - y| < r$. So if $f \in C_b({\bf R}^d)$ and $x \in {\bf R}^d$ then for $R \gg r$ the function $$g(y) = \cases{\big(1 - \frac{|x - y|}{R}\big)f(y)&if |x - y| < R\cr0&if |x - y| \geq R\cr}$$ will satisfy $Ag(y) \approx (1 - \frac{|x-y|}{R})Af(y)$ for $|x - y| < R$. Thus $(\lambda I - A)g(y) \approx (1 - \frac{|x-y|}{R})(\lambda I - A)f(y)$ and the dissipation inequality for $g$ will imply approximately the same inequality for $f$ at $x$.
For general $a$, approximate by $a$ with bounded support. This part is easy.
• Thank you. It's a bit strange, since $C_0(R^d)$ is not dense in $C_b(R^d)$ in $sup$-norm. – user34763 Jun 8 '13 at 4:42
• No, but I guess the point is that dissipativeness of $A$ is a local property, so we don't need uniform approximation, only uniform approximation on compact sets. – Nik Weaver Jun 8 '13 at 5:46
• I used as dissipation inequality $\| (\lambda - A) f\|\geq \lambda \|f\|$. It is not local. You probably mean the equivalent inequality via functionals. But is is true that the corresponding functionals (from Hanh-Banach theorem) will be "the same" for $f$ and $g$? – user34763 Jun 8 '13 at 6:42
• "Local" in the sense that $(\lambda - A)f(x)$ only depends on values of $f$ near $x$, assuming $a$ has bounded support. – Nik Weaver Jun 8 '13 at 7:38