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We consider the concrete Hilbert space $L^{2}(E,m)=L^{2}(E,\mathcal{B},m)$ with usual inner inner product $(\cdot,\cdot)$ where $(E,\mathcal{B},m)$ is a measure space.

For $u,v:E\to \mathbb{R}$,we set \begin{eqnarray*} u\vee v:=\max(u,v),\,u\wedge v:=\min(u,v),\,u^{+}:=u\vee0,\,u^{-}:=-(u\wedge 0) \end{eqnarray*}

$\rm{\underline{Definition}}$ A closed densely defined linear operator $L$ on $L^{2}(E,m)$ is called Dirichlet operator if $(Lu,(u-1)^{+})\leq 0$ for all $u\in D(L)$

I want to show Dirichlet operator is negative definite. i.e. $(Lu,u)\leq 0$ for all $u\in D(L)$.

Since for all $u \in D(L)$, $u=(u-1)^{+}+u\wedge 1$, \begin{eqnarray*} (Lu,u)&=&(Lu,(u-1)^{+}+u\wedge 1)\\ &=&(Lu,(u-1)^{+})+(Lu,u\wedge 1)\\ &\leq&0+(Lu,u\wedge 1)=(Lu,u\wedge1) \end{eqnarray*}

What should I do before that? Please give me a clue.

Sangchul Lee
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ko4
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1 Answers1

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Let $u \in D(L)$ and notice that for $r > 0$, the inequality that you derived also says

$$ (Lu, u) = r^{-2} (L(ru), ru) \leq r^{-2} (L(ru), (ru) \wedge 1) = (Lu, u \wedge r^{-1}). $$

Taking $r \to \infty$, $u \wedge r^{-1} \to -u^{-}$ in $L^{2}$. This gives

$$ (Lu, u) \leq (Lu, -u^{-}) \quad \Longleftrightarrow \quad (Lu, u^{+}) \leq 0. $$

Now replace $u$ by $-u$. Then

$$ (Lu, -u^{-}) = (-Lu, (-u)^{+}) = (L(-u), (-u)^{+}) \leq 0. $$

Adding these two inequalities yields the desired conclusion $(Lu, u) \leq 0$.

Sangchul Lee
  • 167,468