# Composite of Continuous Mappings between Normed Vector Spaces is Continuous

Jump to navigation
Jump to search

## Theorem

Let $X, Y, Z$ be normed vector spaces.

Let $f : X \to Y$ and $g : Y \to Z$ be continuous mappings on $X$ and $Y$ respectively.

Let $g \circ f : X \to Z$ be a composite mapping where:

- $\forall x \in X : \map {\paren {g \circ f} } x := \map g {\map f x}$

Then $g \circ f$ is continuous on $X$.

## Proof

Let $W$ be open in $Z$.

$g$ is continuous on $Y$.

By Mapping is Continuous iff Inverse Images of Open Sets are Open, $\map {g^{-1} } W$ is open in $Y$.

$f$ is continuous on $X$.

By Mapping is Continuous iff Inverse Images of Open Sets are Open, $\map {f^{-1} } {\map {g^{-1} } W}$ is open in $X$.

By Inverse of Composite Relation:

- $\map {\paren {g \circ f}^{-1} } W = \map {f^{-1} } {\map {g^{-1} } W}$

Hence, $\map {\paren {g \circ f}^{-1} } W$ is continuous on $X$.

By Mapping is Continuous iff Inverse Images of Open Sets are Open, $g \circ f$ is continuous on $X$.

$\blacksquare$

## Also see

- Composite of Continuous Mappings is Continuous
- Composite of Continuous Mappings between Metric Spaces is Continuous

## Sources

- 2017: Amol Sasane:
*A Friendly Approach to Functional Analysis*... (previous) ... (next): Chapter $\S 2.2$: Continuous and linear maps. Continuous maps