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bounded operator (Definition)

Definition [1]

  1. Suppose $ X$ and $ Y$ are normed vector spaces with norms $ \Vert \cdot \Vert _X$ and $ \Vert \cdot \Vert _Y$. Further, suppose $ T$ is a linear map $ T : X \to Y$. If there is a $ C \in \mathbf{R}$ such that for all $ x \in X$ we have
    $\displaystyle \Vert Tx \Vert _Y$ $\displaystyle \le$ $\displaystyle C\Vert x \Vert _X,$  

    then $ T$ is a bounded operator.
  2. Let $ X$ and $ Y$ be as above, and let $ T : X \to Y$ be a bounded operator. Then the norm of $ T$ is defined as the real number
    $\displaystyle \Vert T \Vert := \mathrm{sup} \left\{ \left. \frac{\Vert Tx \Vert _Y}{\Vert x \Vert _X} \right\vert x \in X \setminus \{0\} \right\}.$
    Thus the operator norm is the smallest constant $ C \in \mathbf{R}$ such that
    $\displaystyle \Vert Tx \Vert _Y$ $\displaystyle \le$ $\displaystyle C \Vert x \Vert _X.$  

    Now for any $ x \in X \setminus \{0\}$, if we let $ y = x/\Vert x\Vert$, then linearity implies that

    $\displaystyle \Vert Ty \Vert _Y = \left\Vert T\left(\frac{x}{\Vert x \Vert _X}\right) \right\Vert _Y = \frac{\Vert Tx \Vert _Y}{\Vert x\Vert _X}$
    and thus it easily follows that
    $\displaystyle \Vert T \Vert$ $\displaystyle =$ $\displaystyle \mathrm{sup} \left\{ \left. \frac{\Vert Tx \Vert _Y}{\Vert x \Ver... ...Y \left\vert x \in X \setminus \{0\}, y=\frac{x}{\Vert x\Vert} \right. \right\}$  
      $\displaystyle =$ $\displaystyle \mathrm{sup} \{ \Vert Ty \Vert _Y \vert y \in X, \Vert y\Vert = 1 \}.$  

    In the special case when $ X = \{\mathbf{0}\}$ is the zero vector space, any linear map $ T : X \to Y$ is the zero map since $ T(\mathbf{0})=T(0\mathbf{0})=0T(\mathbf{0})=0$. In this case, we define $ \Vert T \Vert := 0$.

  3. To avoid cumbersome notational stuff usually one can simplify the symbols like $ \vert\vert x\vert\vert _X$ and $ \vert\vert Tx\vert\vert _Y$ by writing only $ \vert\vert x\vert\vert$, $ \vert\vert Tx\vert\vert$ since there is a little danger in confusing which is space about calculating norms.

TO DO:

  1. The defined norm for mappings is a norm
  2. Examples: identity operator, zero operator: see [1].
  3. Give alternative expressions for norm of $ T$.
  4. Discuss boundedness and continuity

Theorem [1,2] Suppose $ T : X \to Y$ is a linear map between normed vector spaces $ X$ and $ Y$. If $ X$ is finite-dimensional, then $ T$ is bounded.

Theorem Suppose $ T : X \to Y$ is a linear map between normed vector spaces $ X$ and $ Y$. The following are equivalent:

  1. $ T$ is continuous in some point $ x_{0} \in X$
  2. $ T$ is uniformly continuous in $ X$
  3. $ T$ is bounded

Lemma Any bounded operator with a finite dimensional kernel and cokernel has a closed image.

Proof By Banach's isomorphism theorem.

Bibliography

1
E. Kreyszig, Introductory Functional Analysis With Applications, John Wiley & Sons, 1978.
2
G. Bachman, L. Narici, Functional analysis, Academic Press, 1966.



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Cross-references: isomorphism, image, closed, cokernel, kernel, finite dimensional, uniformly continuous, point, continuous, the following are equivalent, bounded, finite-dimensional, expressions, zero operator, identity operator, mappings, zero map, zero vector space, implies, operator norm, real number, linear map, norms, normed vector spaces
There are 24 references to this entry.

This is version 16 of bounded operator, born on 2003-10-15, modified 2007-08-07.
Object id is 5226, canonical name is BoundedOperator.
Accessed 4058 times total.

Classification:
AMS MSC: 46B99 (Functional analysis :: Normed linear spaces and Banach spaces; Banach lattices :: Miscellaneous)
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