Transpose of unbounded operators between Banach spaces. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T01:33:30Z http://mathoverflow.net/feeds/question/90197 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90197/transpose-of-unbounded-operators-between-banach-spaces Transpose of unbounded operators between Banach spaces. Martin 2012-03-04T13:07:11Z 2012-03-04T19:07:26Z <p>Let $X$ and $Y$ be Banach spaces, and let $L : X \rightarrow Y$ be a unbounded operator with dense domain $\operatorname{dom}(L)$. We can then talk about the transposed operator</p> <p>$L' : \operatorname{dom}(L') \subset Y' \rightarrow X' : y' \rightarrow y'(T\cdot)$</p> <p>whose domain is given by those functionals $y'$, such that the term $y'(T\cdot)$, initially defined on $\operatorname{dom}(L)$, has bounded extension to all of $X$. If $L$ is closed and densely defined, then it is standard to show that $L'$ is closed, too. But if what the density of the domain of transpose? The proof by Reed and Simons seems in the Hilbert space case seems to use specific Hilbert space techniques.</p> <p><strong>Question:</strong> Suppose $L$ is a closed densely-defined operator between Banach spaces. Is it transpose a closed densely-defined operator, too?</p> http://mathoverflow.net/questions/90197/transpose-of-unbounded-operators-between-banach-spaces/90202#90202 Answer by Anatoly Kochubei for Transpose of unbounded operators between Banach spaces. Anatoly Kochubei 2012-03-04T14:00:01Z 2012-03-04T14:00:01Z <p>In general, the transpose need not be densely defined. For an example see</p> <p>S. G. Krein, Linear equations in Banach spaces. Birkhäuser, Boston, 1982.</p> http://mathoverflow.net/questions/90197/transpose-of-unbounded-operators-between-banach-spaces/90203#90203 Answer by Liviu Nicolaescu for Transpose of unbounded operators between Banach spaces. Liviu Nicolaescu 2012-03-04T14:06:37Z 2012-03-04T19:07:26Z <p>The transpose is <strong>closed</strong> but <strong>it may not be densely defined</strong>. For more info see Sec. 2.6 of</p> <blockquote> <p>H. Brezis: Functional Analysis, Sobolev Spaces and Partial Differential Equations, Springer Verlag, 2011</p> </blockquote> <p>Exercise 2.22 in this book describes a closed densely defined operator whose adjoint is not dense. Here it is.</p> <p>Consider the Banach space $E=\ell^1$ with dual $E^*=\ell^\infty$. Consider the densely defined operator</p> <p>$$A: D(A)\subset E\to E,$$</p> <p>$$D(A)=\bigl\lbrace\; (u_n)\in\ell^1;\;\; (nu_n)\in \ell^1 \;\bigr\rbrace, \;\; A(u_n)= (nu_n).$$</p> <p>Then $A$ is closed, densely defined, $A^*$ is closed, but $D(A^*)$ is not dense. </p>