continuous linear mapping

If $(V_1,\parallel \cdot {\parallel }_1)$ and $(V_2,\parallel \cdot {\parallel }_2)$ are normed vector spaces, a linear mapping $T:V_1\to V_2$ is continuous if it is continuous in the metric induced by the norms.

If there is a nonnegative constant $c$ such that${\parallel T(x)\parallel }_2\le c{\parallel x\parallel }_1$ for each $x\in V_1$, we say that $T$ is . This should not be confused with the usual terminology referring to a bounded function as one that has bounded range. In fact, bounded linear mappings usually have unbounded ranges.

The expression bounded linear mapping is often used in functional analysis to refer to continuous linear mappings as well. This is because the two definitions are equivalent:

If $T$ is bounded, then ${\parallel T(x)-T(y)\parallel }_2={\parallel T(x-y)\parallel }_2\le c{\parallel x-y\parallel }_1$, so $T$ is a Lipschitz function. Now suppose $T$ is continuous. Then there exists $r>0$ such that ${\parallel T(x)\parallel }_2\le 1$ when ${\parallel x\parallel }_1\le r$. For any $x\in V_1$, we then have

$$\frac{r}{{\parallel x\parallel }_1}{\parallel T(x)\parallel }_2={\parallel T\left(\frac{r}{{\parallel x\parallel }_1}x\right)\parallel }_2\le 1,$$

hence ${\parallel T(x)\parallel }_2\le r{\parallel x\parallel }_1$; so $T$ is bounded.

It can be shown that a linear mapping between two topological vector spaces is continuous if and only if it is continuous at (http://planetmath.org/Continuous) $0$ [1].