In functional analysis and operator theory, a bounded linear operator is a special kind of linear transformation that is particularly important in infinite dimensions. In finite dimensions, a linear transformation takes a bounded set to another bounded set (for example, a rectangle in the plane goes either to a parallelogram or bounded line segment when a linear transformation is applied). However, in infinite dimensions, linearity is not enough to ensure that bounded sets remain bounded: a bounded linear operator is thus a linear transformation that sends bounded sets to bounded sets.

Formally, it is a linear transformation ${\displaystyle L:X\to Y}$ between topological vector spaces (TVSs) ${\displaystyle X}$ and ${\displaystyle Y}$ that maps bounded subsets of ${\displaystyle X}$ to bounded subsets of ${\displaystyle Y.}$ If ${\displaystyle X}$ and ${\displaystyle Y}$ are normed vector spaces (a special type of TVS), then ${\displaystyle L}$ is bounded if and only if there exists some ${\displaystyle M>0}$ such that for all ${\displaystyle x\in X,}$ $${\displaystyle \|Lx\|_{Y}\leq M\|x\|_{X}.}$$ The smallest such ${\displaystyle M}$ is called the operator norm of ${\displaystyle L}$ and denoted by ${\displaystyle \|L\|.}$ A linear operator between normed spaces is continuous if and only if it is bounded.

The concept of a bounded linear operator has been extended from normed spaces to all topological vector spaces.

Outside of functional analysis, when a function ${\displaystyle f:X\to Y}$ is called "bounded" then this usually means that its image ${\displaystyle f(X)}$ is a bounded subset of its codomain. A linear map has this property if and only if it is identically ${\displaystyle 0.}$ Consequently, in functional analysis, when a linear operator is called "bounded" then it is never meant in this abstract sense (of having a bounded image).

Every bounded operator is Lipschitz continuous at ${\displaystyle 0.}$

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

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 $${\displaystyle \|Lx\|=\left\Vert {\|x\| \over \varepsilon }L\left(\varepsilon {x \over \|x\|}\right)\right\Vert ={\|x\| \over \varepsilon }\left\Vert L\left(\varepsilon {x \over \|x\|}\right)\right\Vert \leq {\|x\| \over \varepsilon }\cdot 1={1 \over \varepsilon }\|x\|.}$$ This proves that ${\displaystyle L}$ is bounded. Q.E.D.