https://en.formulasearchengine.com/api.php?action=feedcontributions&user=2001%3A6A0%3A5001%3A2%3A6171%3A3BB7%3AA19D%3A7717&feedformat=atom formulasearchengine - User contributions [en] 2021-09-27T16:32:37Z User contributions MediaWiki 1.37.0-alpha https://en.formulasearchengine.com/index.php?title=Bounded_operator&diff=230693 Bounded operator 2012-07-24T09:56:41Z <p>2001:6A0:5001:2:6171:3BB7:A19D:7717: generally it was not true! Rudin 1.32</p> <hr /> <div>{{Unreferenced|date=December 2009}}<br /> In [[functional analysis]], a branch of [[mathematics]], a '''bounded linear operator''' is a [[linear transformation]] ''L'' between [[normed vector space]]s ''X'' and ''Y'' for which the ratio of the norm of ''L''(''v'') to that of ''v'' is [[bounded set|bounded]] by the same number, over all non-zero vectors ''v'' in ''X''. In other words, there exists some ''M''&amp;nbsp;&gt;&amp;nbsp;0 such that for all ''v'' in ''X'' <br /> :&lt;math&gt;\|Lv\|_Y \le M \|v\|_X.\, \,&lt;/math&gt;<br /> <br /> The smallest such ''M'' is called the [[operator norm]] &lt;math&gt;\|L\|_{\mathrm{op}} \,&lt;/math&gt; of ''L''.<br /> <br /> A bounded linear operator is generally not a [[bounded function]]; the latter would require that the norm of ''L''(''v'') be bounded for all ''v'', which is not possible unless ''Y'' is the zero vector space. Rather, a bounded linear operator is a [[locally bounded function]]. <br /> <br /> A linear operator on a metrizable vector space is bounded if and only if it is [[continuous linear operator|continuous]].<br /> <br /> ==Examples==<br /> * Any linear operator between two finite-dimensional normed spaces is bounded, and such an operator may be viewed as multiplication by some fixed [[matrix (mathematics)|matrix]].<br /> <br /> * Many [[integral transform]]s are bounded linear operators. For instance, if <br /> ::&lt;math&gt;K:[a, b]\times [c, d]\to {\mathbf R} \,&lt;/math&gt;<br /> :is a continuous function, then the operator &lt;math&gt;L, \,&lt;/math&gt; defined on the space &lt;math&gt;C[a, b] \,&lt;/math&gt; of continuous functions on &lt;math&gt;[a, b] \,&lt;/math&gt; endowed with the [[uniform norm]] and with values in the space &lt;math&gt;C[c, d], \,&lt;/math&gt; with &lt;math&gt;L \,&lt;/math&gt; given by the formula<br /> ::&lt;math&gt;(Lf)(y)=\int_{a}^{b}\!K(x, y)f(x)\,dx, \,&lt;/math&gt;<br /> :is bounded. This operator is in fact [[compact operator|compact]]. The compact operators form an important class of bounded operators. <br /> <br /> * The [[Laplace operator]] <br /> ::&lt;math&gt;\Delta:H^2({\mathbf R}^n)\to L^2({\mathbf R}^n) \,&lt;/math&gt;<br /> :(its domain is a [[Sobolev space]] and it takes values in a space of [[square integrable function]]s) is bounded.<br /> <br /> * The [[shift operator]] on the [[Lp space|''l''&lt;sup&gt;''2''&lt;/sup&gt; space]] of all [[sequence]]s (''x''&lt;sub&gt;''0''&lt;/sub&gt;, ''x''&lt;sub&gt;''1''&lt;/sub&gt;, ''x''&lt;sub&gt;''2''&lt;/sub&gt;...) of real numbers with &lt;math&gt;x_0^2+x_1^2+x_2^2+\cdots &lt; \infty, \,&lt;/math&gt; <br /> ::&lt;math&gt;L(x_0, x_1, x_2, \dots)=(0, x_0, x_1, x_2,\dots) \,&lt;/math&gt; <br /> :is bounded. Its operator norm is easily seen to be 1.<br /> <br /> ==Equivalence of boundedness and continuity==<br /> As stated in the introduction, a linear operator ''L'' between normed spaces ''X'' and ''Y'' is bounded if and only if it is a [[continuous linear operator]]. The proof is as follows.<br /> <br /> * Suppose that ''L'' is bounded. Then, for all vectors ''v'' and ''h'' in ''X'' with ''h'' nonzero we have<br /> ::&lt;math&gt;\|L(v + h) - L v\| = \|Lh\| \le M\|h\|. \,&lt;/math&gt;<br /> :Letting &lt;math&gt;\mathit{h} \,&lt;/math&gt; go to zero shows that ''L'' is continuous at ''v''. Moreover, since the constant ''M'' does not depend on ''v'', this shows that in fact ''L'' is [[uniform continuity|uniformly continuous]] (Even stronger, it is [[Lipschitz continuous]].)<br /> <br /> * Conversely, it follows from the continuity at the zero vector that there exists a &lt;math&gt;\delta &gt; 0&lt;/math&gt; such that &lt;math&gt;\|L(h)\|=\| L(h) - L(0) \| \le 1&lt;/math&gt; for all vectors ''h'' in ''X'' with &lt;math&gt;\|h\| \le \delta&lt;/math&gt;. Thus, for all non-zero &lt;math&gt;v&lt;/math&gt; in ''X'', one has<br /> ::&lt;math&gt;\|Lv\| = \left \Vert {\|v\| \over \delta} L \left( \delta {v \over \|v\|} \right) \right \Vert = {\|v\| \over \delta} \left \Vert L \left( \delta {v \over \|v\|} \right) \right \Vert \le {\|v\| \over \delta} \cdot 1 = {1 \over \delta}\|v\|. &lt;/math&gt;<br /> : This proves that ''L'' is bounded.<br /> <br /> ==Linearity and boundedness==<br /> Not every linear operator between normed spaces is bounded. Let ''X'' be the space of all [[trigonometric polynomial]]s defined on [&amp;minus;π, π], with the norm <br /> :&lt;math&gt;\|P\|=\int_{-\pi}^{\pi}\!|P(x)|\,dx.&lt;/math&gt;<br /> Define the operator ''L'':''X''→''X'' which acts by taking the [[derivative]], so it maps a polynomial ''P'' to its derivative ''P''&amp;prime;. Then, for <br /> :&lt;math&gt;v=e^{in x}&lt;/math&gt;<br /> with ''n''=1, 2, ...., we have &lt;math&gt;\|v\|=2\pi,&lt;/math&gt; while &lt;math&gt;\|L (v)\|=2\pi n\to\infty&lt;/math&gt; as ''n''→∞, so this operator is not bounded.<br /> <br /> It turns out that this is not a singular example, but rather part of a general rule. Any linear operator defined on a finite-dimensional normed space is bounded. However, given any normed spaces ''X'' and ''Y'' with ''X'' infinite-dimensional and ''Y'' not being the zero space, one can find a [[discontinuous linear map|linear operator which is not continuous]] from ''X'' to ''Y''.<br /> <br /> That such a basic operator as the derivative (and others) is not bounded makes it harder to study. If, however, one defines carefully the domain and range of the derivative operator, one may show that it is a [[closed operator]]. Closed operators are more general than bounded operators but still &quot;well-behaved&quot; in many ways.<br /> <br /> ==Further properties==<br /> The condition for ''L'' to be bounded, namely that there exists some ''M'' such that for all ''v''<br /> :&lt;math&gt;\|Lv\| \le M \|v\|,\,&lt;/math&gt;<br /> is precisely the condition for ''L'' to be [[Lipschitz continuous]] at 0 (and hence, everywhere, because ''L'' is linear).<br /> <br /> A common procedure for defining a bounded linear operator between two given [[Banach space|Banach]] spaces is as follows. First, define a linear operator on a [[dense set|dense subset]] of its domain, such that it is locally bounded. Then, extend the operator by continuity to a continuous linear operator on the whole domain (see [[continuous linear extension]]).<br /> <br /> ==Properties of the space of bounded linear operators==<br /> <br /> * The space of all bounded linear operators from ''U'' to ''V'' is denoted by ''B''(''U'',''V'') and is a normed vector space.<br /> * If ''V'' is Banach, then so is ''B''(''U'',''V''),<br /> * from which it follows that [[dual space]]s are Banach.<br /> * For any ''A'' in ''B''(''U'',''V''), the kernel of ''A'' is a closed linear subspace of ''U''.<br /> * If ''B''(''U'',''V'') is Banach and ''U'' is nontrivial, then ''V'' is Banach.<br /> <br /> ==Topological vector spaces==<br /> The boundedness condition for linear operators on normed spaces can be restated. An operator is bounded if it takes every [[bounded set (topological vector space)|bounded set]] to a bounded set, and here is meant the more general condition of boundedness for sets in a [[topological vector space]] (TVS): a set is bounded if and only if it is absorbed by every neighborhood of 0. Note that the two notions of boundedness coincide for [[locally convex spaces]].<br /> <br /> This formulation allows one to define bounded operators between general topological vector spaces as an operator which takes bounded sets to bounded sets. In this context, it is still true that every continuous map is bounded, however the converse fails; a bounded operator need not be continuous. Clearly, this also means that boundedness is no longer equivalent to Lipschitz continuity in this context.<br /> <br /> A converse does hold when the domain is pseudometrisable, a case which includes [[Fréchet space]]s. For [[LF space]]s, a weaker converse holds; any bounded linear map from an LF space is [[sequentially continuous]].<br /> <br /> ==See also==<br /> *[[Operator algebra]]<br /> *[[Operator theory]]<br /> *[[Unbounded operator]]<br /> <br /> ==References==<br /> * Kreyszig, Erwin: ''Introductory Functional Analysis with Applications'', Wiley, 1989<br /> <br /> {{DEFAULTSORT:Bounded Operator}}<br /> [[Category:Linear operators]]<br /> [[Category:Operator theory]]<br /> [[Category:Continuous mappings]]<br /> <br /> [[es:Operador lineal acotado]]<br /> [[fr:Opérateur borné]]<br /> [[it:Operatore limitato]]<br /> [[he:אופרטור לינארי חסום]]<br /> [[ja:有界作用素]]<br /> [[pl:Operator liniowy ograniczony]]<br /> [[pt:Operador linear limitado]]<br /> [[ru:Ограниченный оператор]]<br /> [[zh:有界算符]]</div> 2001:6A0:5001:2:6171:3BB7:A19D:7717