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Let $V, H$ two Hilbert spaces infinite dimensional.

If the bilinear form $a(.,.)$ satisfies

  1. There exists a constant $\alpha>0$ such that $\forall v \in V, \ a(v,v)\geq \alpha \|v\|^2.$
  2. There exists a constant $M>0$ such that $\forall u, v \in V, \ a(u,v)\leq M \|u\| \|v\|.$

it is possible to definite the operator $T$ for which $$\forall v \in V, \ a(Tu,v)=(u,v)$$ This is a conseguence of Lax-Milgram theorem, if $u\in H$ and if $V\subset H$ compactly and continuously embedded, the foregoing problem admits a unique solution $Tu\in V$. T is linear from $V$ in $H$. Not only, thanks to the fact that $V\subset H$ compactly and continuously embedded, the operator $T$ is compact from $V$ in $V$.

Under these assumptions, answer the following questions: Is $T$ a bounded operator? Is it a regular operator?

I have answered the first question. The answer is: yes. In fact, if $V\subset H$ compactly and continuously embedded, there exists a constant $c>0$ such that $$\forall v\in V, \ |v|\leq c\|v\|$$ where $|.|$ is the norm on H, and $\|.\|$ is the norm on V. By assumption $(1)$: $$\|Tu\|\leq \frac{1}{\alpha} \sup_{v\in V} \frac{(u,v)}{\|v\|}\leq \frac{c}{\alpha} |u|$$

Any suggestions to answer the second question, please?

Thank you very much.

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  • $\begingroup$ It would be much easier if you distinguish between the norms/scalar products in $V$ and $H$. $\endgroup$
    – gerw
    Jun 9, 2015 at 10:25
  • $\begingroup$ Really? In what way? $\endgroup$
    – Mark
    Jun 9, 2015 at 10:39
  • $\begingroup$ What do you mean by 'regular operator'? As $T$ is compact from $V$ to $V$ it cannot be continuously invertible (if that is what you are after). $\endgroup$
    – daw
    Jun 10, 2015 at 13:54
  • $\begingroup$ There is some conditions for the regularity of a bounded operator $L$; for example, I know the representability of $L$ as $L = I + T$, where $I$ is the unit, and $T$ a completely continuous operator for which $- 1$ is not an eigenvalue. However, your comment has answered my question. If $T$ isn't continuously invertible then it cannot be regular. Thanks. $\endgroup$
    – Mark
    Jun 10, 2015 at 19:47

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