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− | Linear operators <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c0234501.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c0234502.png" />, of which <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c0234503.png" /> is of general type and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c0234504.png" /> is bounded, and which are such that | + | {{TEX|done}} |
| + | Linear operators $B$ and $T$, of which $T$ is of general type and $B$ is bounded, and which are such that |
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− | <table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c0234505.png" /></td> <td valign="top" style="width:5%;text-align:right;">(1)</td></tr></table>
| + | $$BT\subseteq TB\tag{1}$$ |
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− | (the symbol <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c0234506.png" /> means that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c0234507.png" /> is an extension of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c0234508.png" />, cf. [[Extension of an operator|Extension of an operator]]). The commutation relation is denoted by <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c0234509.png" /> and satisfies the following rules: | + | (the symbol $T\subseteq T_1$ means that $T_1$ is an extension of $T$, cf. [[Extension of an operator|Extension of an operator]]). The commutation relation is denoted by $B\cup T$ and satisfies the following rules: |
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− | 1) if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345010.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345011.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345012.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345013.png" />; | + | 1) if $B\cup T_1$, $B\cup T_2$, then $B\cup(T_1+T_2)$, $B\cup T_1T_2$; |
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− | 2) if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345014.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345015.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345016.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345017.png" />; | + | 2) if $B_1\cup T$, $B_2\cup T$, then $(B_1+B_2)\cup T$, $B_1B_2\cup T$; |
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− | 3) if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345018.png" /> exists, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345019.png" /> implies that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345020.png" />; | + | 3) if $T^{-1}$ exists, then $B\cup T$ implies that $B\cup T^{-1}$; |
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− | 4) if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345021.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345022.png" /> then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345023.png" />; | + | 4) if $B\cup T_n$, $n=1,2,\dots,$ then $B\cup\lim T_n$; |
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− | 5) if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345024.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345025.png" /> then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345026.png" />, provided that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345027.png" /> is bounded and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345028.png" /> is closed. | + | 5) if $B_n\cup T$, $n=1,2,\dots,$ then $\lim B_n\cup T$, provided that $\lim B_n$ is bounded and $T$ is closed. |
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| If the two operators are defined on the entire space, condition 1) reduces to the usual one: | | If the two operators are defined on the entire space, condition 1) reduces to the usual one: |
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− | <table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345029.png" /></td> <td valign="top" style="width:5%;text-align:right;">(2)</td></tr></table>
| + | $$BT=TB,\tag{2}$$ |
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− | and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345030.png" /> is not required to be bounded. The generalization of (2) is justified by the fact that even a bounded operator <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345031.png" /> need not commute with its inverse <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023450/c02345032.png" /> if the latter is not defined on the entire space. | + | and $B$ is not required to be bounded. The generalization of \ref{2} is justified by the fact that even a bounded operator $B$ need not commute with its inverse $B^{-1}$ if the latter is not defined on the entire space. |
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| ====References==== | | ====References==== |
| <table><TR><TD valign="top">[1]</TD> <TD valign="top"> L.A. Lyusternik, V.I. Sobolev, "Elements of functional analysis" , Hindushtan Publ. Comp. (1974) (Translated from Russian)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top"> F. Riesz, B. Szökefalvi-Nagy, "Functional analysis" , F. Ungar (1955) (Translated from French)</TD></TR></table> | | <table><TR><TD valign="top">[1]</TD> <TD valign="top"> L.A. Lyusternik, V.I. Sobolev, "Elements of functional analysis" , Hindushtan Publ. Comp. (1974) (Translated from Russian)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top"> F. Riesz, B. Szökefalvi-Nagy, "Functional analysis" , F. Ungar (1955) (Translated from French)</TD></TR></table> |
Linear operators $B$ and $T$, of which $T$ is of general type and $B$ is bounded, and which are such that
$$BT\subseteq TB\tag{1}$$
(the symbol $T\subseteq T_1$ means that $T_1$ is an extension of $T$, cf. Extension of an operator). The commutation relation is denoted by $B\cup T$ and satisfies the following rules:
1) if $B\cup T_1$, $B\cup T_2$, then $B\cup(T_1+T_2)$, $B\cup T_1T_2$;
2) if $B_1\cup T$, $B_2\cup T$, then $(B_1+B_2)\cup T$, $B_1B_2\cup T$;
3) if $T^{-1}$ exists, then $B\cup T$ implies that $B\cup T^{-1}$;
4) if $B\cup T_n$, $n=1,2,\dots,$ then $B\cup\lim T_n$;
5) if $B_n\cup T$, $n=1,2,\dots,$ then $\lim B_n\cup T$, provided that $\lim B_n$ is bounded and $T$ is closed.
If the two operators are defined on the entire space, condition 1) reduces to the usual one:
$$BT=TB,\tag{2}$$
and $B$ is not required to be bounded. The generalization of \ref{2} is justified by the fact that even a bounded operator $B$ need not commute with its inverse $B^{-1}$ if the latter is not defined on the entire space.
References
[1] | L.A. Lyusternik, V.I. Sobolev, "Elements of functional analysis" , Hindushtan Publ. Comp. (1974) (Translated from Russian) |
[2] | F. Riesz, B. Szökefalvi-Nagy, "Functional analysis" , F. Ungar (1955) (Translated from French) |