Bounded operator

A linear operator between normed spaces is bounded if and only if it is continuous.

Proof

Suppose that ${\displaystyle L}$ is bounded. Then, for all vectors ${\displaystyle x,h\in X}$ with ${\displaystyle h}$ nonzero we have

$${\displaystyle \|L(x+h)-L(x)\|=\|L(h)\|\leq M\|h\|.}$$

Letting ${\displaystyle h}$ go to zero shows that ${\displaystyle L}$ is continuous at ${\displaystyle x.}$ Moreover, since the constant ${\displaystyle M}$ does not depend on ${\displaystyle x,}$ this shows that in fact ${\displaystyle L}$ is uniformly continuous, and even Lipschitz continuous.

Conversely, it follows from the continuity at the zero vector that there exists a ${\displaystyle \varepsilon >0}$ such that ${\displaystyle \|L(h)\|=\|L(h)-L(0)\|\leq 1}$ for all vectors ${\displaystyle h\in X}$ with ${\displaystyle \|h\|\leq \varepsilon .}$ Thus, for all non-zero ${\displaystyle x\in X,}$ one has

$${\displaystyle \|Lx\|=\left\Vert {\|x\| \over \varepsilon }L\left(\varepsilon {x \over \|x\|}\right)\right\Vert ={\|x\| \over \varepsilon }\left\Vert L\left(\varepsilon {x \over \|x\|}\right)\right\Vert \leq {\|x\| \over \epsilon }\cdot 1={1 \over \varepsilon }\|x\|.}$$

This proves that ${\displaystyle L}$ is bounded. Q.E.D.

Every sequentially continuous linear operator between TVS is a bounded operator.[1] This implies that every continuous linear operator between metrizable TVS is bounded. However, in general, a bounded linear operator between two TVSs need not be continuous.

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. This also means that boundedness is no longer equivalent to Lipschitz continuity in this context.

If the domain is a bornological space (for example, a pseudometrizable TVS, a Fréchet space, a normed space) then a linear operators into any other locally convex spaces is bounded if and only if it is continuous. For LF spaces, a weaker converse holds; any bounded linear map from an LF space is sequentially continuous.

If ${\displaystyle F:X\to Y}$ is a linear operator between two topological vector spaces and if there exists a neighborhood ${\displaystyle U}$ of the origin in ${\displaystyle X}$ such that ${\displaystyle F(U)}$ is a bounded subset of ${\displaystyle Y,}$ then ${\displaystyle F}$ is continuous.[2] This fact is often summarized by saying that a linear operator that is bounded on some neighborhood of the origin is necessarily continuous. In particular, any linear functional that is bounded on some neighborhood of the origin is continuous (even if its domain is not a normed space).

Bornological spaces are exactly those locally convex spaces for which every bounded linear operator into another locally convex space is necessarily continuous. That is, a locally convex TVS ${\displaystyle X}$ is a bornological space if and only if for every locally convex TVS ${\displaystyle Y,}$ a linear operator ${\displaystyle F:X\to Y}$ is continuous if and only if it is bounded.[3]

Every normed space is bornological.

Let ${\displaystyle F:X\to Y}$ be a linear operator between topological vector spaces (not necessarily Hausdorff). The following are equivalent:

1. ${\displaystyle F}$ is (locally) bounded;[3]
2. (Definition): ${\displaystyle F}$ maps bounded subsets of its domain to bounded subsets of its codomain;[3]
3. ${\displaystyle F}$ maps bounded subsets of its domain to bounded subsets of its image ${\displaystyle \operatorname {Im} F:=F(X)}$;[3]
4. ${\displaystyle F}$ maps every null sequence to a bounded sequence;[3]
   • A null sequence is by definition a sequence that converges to the origin.
   • Thus any linear map that is sequentially continuous at the origin is necessarily a bounded linear map.
5. ${\displaystyle F}$ maps every Mackey convergent null sequence to a bounded subset of ${\displaystyle Y.}$[note 1]
   • A sequence ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }}$ is said to be Mackey convergent to the origin in ${\displaystyle X}$ if there exists a divergent sequence ${\displaystyle r_{\bullet }=\left(r_{i}\right)_{i=1}^{\infty }\to \infty }$ of positive real number such that ${\displaystyle r_{\bullet }=\left(r_{i}x_{i}\right)_{i=1}^{\infty }}$ is a bounded subset of ${\displaystyle X.}$

if ${\displaystyle X}$ and ${\displaystyle Y}$ are locally convex then the following may be add to this list:

6. ${\displaystyle F}$ maps bounded disks into bounded disks.[4]
7. ${\displaystyle F^{-1}}$ maps bornivorous disks in ${\displaystyle Y}$ into bornivorous disks in ${\displaystyle X.}$[4]

if ${\displaystyle X}$ is a bornological space and ${\displaystyle Y}$ is locally convex then the following may be added to this list:

8. ${\displaystyle F}$ is sequentially continuous at some (or equivalently, at every) point of its domain.[5]
   • A sequentially continuous linear map between two TVSs is always bounded,[1] but the converse requires additional assumptions to hold (such as the domain being bornological and the codomain being locally convex).
   • If the domain ${\displaystyle X}$ is also a sequential space, then ${\displaystyle F}$ is sequentially continuous if and only if it is continuous.
9. ${\displaystyle F}$ is sequentially continuous at the origin.

Let ${\displaystyle X}$ be the space of all trigonometric polynomials on ${\displaystyle [-\pi ,\pi ],}$ with the norm

$${\displaystyle \|P\|=\int _{-\pi }^{\pi }\!|P(x)|\,dx.}$$

The operator ${\displaystyle L:X\to X}$ that maps a polynomial to its derivative is not bounded. Indeed, for ${\displaystyle v_{n}=e^{inx}}$ with ${\displaystyle n=1,2,\ldots ,}$ we have ${\displaystyle \|v_{n}\|=2\pi ,}$ while ${\displaystyle \|L(v_{n})\|=2\pi n\to \infty }$ as ${\displaystyle n\to \infty ,}$ so ${\displaystyle L}$ is not bounded.

• The space of all bounded linear operators from ${\displaystyle X}$ to ${\displaystyle Y}$ is denoted by ${\displaystyle B(X,Y)}$ and is a normed vector space.
• If ${\displaystyle Y}$ is Banach, then so is ${\displaystyle B(X,Y).}$
• from which it follows that dual spaces are Banach.
• For any ${\displaystyle A\in B(X,Y),}$ the kernel of ${\displaystyle A}$ is a closed linear subspace of ${\displaystyle X.}$
• If ${\displaystyle B(X,Y)}$ is Banach and ${\displaystyle X}$ is nontrivial, then ${\displaystyle Y}$ is Banach.