Editing Bounded operator (section)

==In normed vector spaces==

Every bounded operator is [[Lipschitz continuity|Lipschitz continuous]] at <math>0.</math>

===Equivalence of boundedness and continuity===

A linear operator between normed spaces is bounded if and only if it is [[Continuous linear operator|continuous]].

{{math proof|title=Proof|proof=
Suppose that <math>L</math> is bounded. Then, for all vectors <math>x, h \in X</math> with <math>h</math> nonzero we have
<math display=block>\|L(x + h) - L(x)\| = \|L(h)\| \leq M\|h\|.</math>
Letting <math>h </math> go to zero shows that <math>L</math> is continuous at <math>x.</math> 
Moreover, since the constant <math>M</math> does not depend on <math>x,</math> this shows that in fact <math>L</math> is [[Uniform continuity|uniformly continuous]], and even [[Lipschitz continuous]].

Conversely, it follows from the continuity at the zero vector that there exists a <math>\varepsilon > 0</math> such that <math>\|L(h)\| = \|L(h) - L(0)\| \leq 1</math> for all vectors <math>h \in X</math> with <math>\|h\| \leq \varepsilon.</math> 
Thus, for all non-zero <math>x \in X,</math> one has
<math display=block>\|Lx\| = \left\Vert {\|x\| \over \varepsilon} L\left(\varepsilon {x \over \|x\|}\right) \right\Vert = {\|x\| \over \varepsilon}\left\Vert L\left(\varepsilon {x \over \|x\|}\right) \right\Vert \leq {\|x\| \over \varepsilon} \cdot 1  = {1 \over \varepsilon}\|x\|.</math>
This proves that <math>L</math> is bounded. [[Q.E.D.]]
}}