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In the paper [1] the following $C_0$-group is presented, $$ T(t)f(x) = f(e^{-t} x) , \quad x \in (0,\infty) \quad f \in E $$ where $E$ is an ($L^1,L^\infty$)-interpolation space. In mi case, I'm just interested in the case $L^p(0,\infty)$ with $1\leq p < \infty$.

In the paper they mention, but they don't prove it, that the infinitesimal generator $A$ is given by $$ Af(x) = -xf'(x), \,\, a.e. \,\,\,\, x \in (0,\infty) $$ with domain $$ D(A) = \{f \in L^p(0,\infty) : f \in AC_{loc}(0,\infty) \text{ and } xf'(x) \in L^p(0,\infty)\}. $$

My attempt

Let $f \in D(A)$, then $g := \lim_{t\to 0} \frac{T(t)f-f}{t} \in L^p(0,\infty)$. As it converges in the $L^p$ norm there exists a subsequence where the convergence is pointwise and therefore we have that,

$$ \lim_{k\to 0} \frac{T(t_k)f(x)-f(x)}{t} = \lim_{k\to 0} \frac{f(e^{-t_k}x) - f(x)}{e^{-t_k} x-x} \frac{e^{-t_k}x-x}{t_k} = -x f'(x), $$ almost everywhere with $x \in (0,\infty)$. Thus, $x f'(x) \in L^p(0,\infty)$.

Now, I'm stuck into proving that $f \in AC_{loc}(0,\infty)$, what I have done so far is noticing that as $xf'(x) \in L^p(0,\infty)$ we have that $(xf(x))' \in L^p(0,\infty)$ and this derivative exists almost everywhere but I don't know really how to follow as I don't know much properties about locally absolutelt continuous functions. Additionally, I'm curious whether the space $AC_{loc}$ could be replaced with $AC$ without loss of generality.

References

[1] Arendt, W., & Pagter, B. (2002). Spectrum and asymptotics of the Black-Scholes partial differential equation in $(L^1,L^\infty)$-interpolation spaces. Pacific J. Math., 202(1), 1–36.

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  • $\begingroup$ No, local absolute continuity must not be replaced with the global version. For example, if $f=\log x$ near $x=0$, then $f\in D(A)$, but $f$ is not globally AC. (There is a similar issue near $\infty$.) $\endgroup$
    – Christian Remling
    Commented May 1 at 20:56
  • $\begingroup$ I think you can argue in the same style as in my answer here (there are probably many arguments and quite possible better ones): mathoverflow.net/questions/469466/… If $(T_h f -f)/h\to g$ in $L^p$, then this convergence also takes place in $\mathcal D'$, which should imply that $f'=g$ in $\mathcal D'$, with $g\in L^p\subseteq L^1_{\textrm{loc}}$, so $f\in AC$. $\endgroup$
    – Christian Remling
    Commented May 1 at 21:03
  • $\begingroup$ You might find the MSE-Q math.stackexchange.com/questions/116633/… (and the links in the body and comments therein) interesting. The combinatorics associated with $x^2D_x$ and $xD_x$ are well studied. $\endgroup$
    – Tom Copeland
    Commented May 3 at 0:24

1 Answer 1

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One proof uses basic tools of semigroup theory (as in Pazy or Engel-Nagel). $T_t$ is strongly continuous in $L^p$ and let $(B,D)$ its generator. One easily checks that $C_c^\infty \subset D$, with $C_c^\infty$ denoting all smooth functions with compact support in $(0, \infty)$, and $Bf=-xf'$. Also, $C_c^\infty$ is $T_t$ invariant, hence a core for $(B,D)$, by the core theorem, which means that it is dense in $D$ for the graph norm. However, it is also dense in $D(A)$ by usual arguments (first cut-off near zero and then mollify) so that $A=B$.

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  • $\begingroup$ Okay I think I understood the first part, but I'm struggling into proving that $C_c^\infty$ is dense in $D(A)$. The idea is to take a function $f \in D(A)$ and try to find a sequence $\{f_n\} \subset C_c^\infty(0,\infty)$. I might take $f_n = \chi_{[1/n,n]} f $ which is of compact support but, how can I prove it's infinite differentiable? Because I know f is $AC$ in [1/n,n] so I can differentiate once but not infinite times. Thank you! $\endgroup$
    – Scottish Questions
    Commented May 2 at 8:56
  • $\begingroup$ First cut-off with $\phi_n(x)=\phi (nx)$ with a suitable smooth $\phi$. Then, with a fixed $n$, use a convolution. That is the idea... $\endgroup$
    – Giorgio Metafune
    Commented May 2 at 9:24
  • $\begingroup$ Okay I see that the idea is to prove it as when you prove $C_c^\infty$ is dense in $L^p$. The thing that I don't really understand is the role of $AC_{loc}$ because the way I see it we can define $D(A) = \{f \in L^p(0,1) \text{ and } xf'(x) \in L^p(0,1)\}$ and we still have that $C_c^\infty$ is dense in $D(A)$ with the graph norm, right? $\endgroup$
    – Scottish Questions
    Commented May 2 at 13:15
  • $\begingroup$ It is easier to use to use $\{ f \in L^p, f \in W^{1,p}_{loc}, xf' \in L^p\}$. It is the same, but the use of Sobolev spaces simplifies my understanding. The norm is $\|f\|_p+\|xf'\|_p$. First one sees that $\phi_n f \to f$ in this norm (the presence of $x$ is crucial to kill the n in front of the derivative). Once $n$ is fixed, $\phi_n f'$ has compact support away from 0 and you can use convolutions, forgettting the $x$. $\endgroup$
    – Giorgio Metafune
    Commented May 2 at 17:05
  • $\begingroup$ Alright, I think I finally got it, seen it with the $W_{loc}^{1,p}$ space makes everything easier for me. Nonetheless, I don't know where comes the intuition to think about the space locally ($W_{loc}^{1,p}$) instead of globally. But I think is an issue about density. $\endgroup$
    – Scottish Questions
    Commented May 7 at 8:30

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