# Operator satisfying $\langle Pu, \phi\rangle=\langle u, ^tP\phi\rangle$?

Let $\Omega\subseteq \mathbb R^n$ be an open subset and $P:\mathscr{D}^\prime(\Omega)\longrightarrow \mathscr{D}^\prime(\Omega)$ be a continuous linear operator. Is it true that there exists only one continuous linear operator $^tP:C^\infty_0(\Omega)\longrightarrow C^\infty_0(\Omega)$ such that $$\langle Pu, \phi\rangle=\langle u, ^tP\phi\rangle?$$

Thanks

The unicity is quite easy to prove: suppose that you have two endomorphisms of $D(\Omega)$ with the property $$\forall u\in D'(\Omega),\,\forall \phi\in D(\Omega)\quad \langle P[u],\phi\rangle=\langle u,A[\phi]\rangle=\langle u,B[\phi]\rangle.$$

This implies $$\forall u\in D'(\Omega),\,\forall \phi\in D(\Omega)\quad \langle u,(A-B)[\phi]\rangle=0.$$

The function $(A-B)[\phi]$ is a test function. If the operator $A-B\ne 0$, then for at least one test function $\phi$ the test function $(A-B)[\phi]$ has a non-empty Lebesgue set $E_0=\{x\in \Omega:\,(A-B)[\phi](x)>0\}$. Take now $u=\mathbf 1_{x\in E_0}(x)$. This ensures us that $\langle u,(A-B)[\phi]\rangle>0$.

Therefore, $\forall \phi \quad (A-B)[\phi]=0$, or, in other words, $A=B$.

The existence of such an operator $A$ ($^tP$ in your notations), however, is a trickier thing to prove.

• Thanks for your answer. Could you recommend me a reference for the proof of the existence?
– PtF
Commented Oct 26, 2015 at 15:25
• @PtF you are welcome. Unfortunately, I don't know such a reference. If you find one - ping me here, I'm interested, too. Commented Oct 26, 2015 at 15:35