Linear Operators Between Vector Spaces

# Linear Operators Between Vector Spaces

## Linear Operators / Transformations

Definition: Let $E$ and $F$ be vector spaces over $\mathbf{F}$. A mapping $f : E \to F$ is a Linear Operator (or Linear Transformation) if it satisfies the following properties:(1) $f(x + y) = f(x) + f(y)$ for all $x, y \in E$.(2) $f(\lambda x) = \lambda f(x)$ for all $x \in E$ and for all $\lambda \in \mathbf{F}$.The Set of All Linear Operators from $E$ to $F$ is denoted by $\mathcal L (E, F)$, and is itself a vector space with the addition and scalar multiplication respectively defined by $(f + g)(x) := f(x) + g(x)$ and $(\lambda f)(x) := \lambda f(x)$ for all $f, g \in \mathcal L(E, F)$ and for all $\lambda \in \mathbf{F}$. |

## Continuity of Linear Operators

When both $E$ and $F$ are topological vector spaces, we can discuss what it means for linear operators from $E$ to $F$ to be continuous.

Proposition 1: Let $E$ and $F$ be topological vector spaces over $\mathbf{F}$. Then, a linear operator $f : E \to F$ is continuous if and only if $f$ is continuous at the origin in $E$. |

**Proof:**$\Rightarrow$ Clearly if $f : E \to F$ is continuous then $f$ is continuous at the origin in $E$.

- $\Leftarrow$ Let $e \in E$. Since $F$ is a topological vector space, every neighbourhood of $f(e)$ is of the form $V + f(e)$ where $V$ is a neighbourhood of the origin. So let $V + f(e)$ be a neighbourhood of $f(e)$ with $V$ a neighbourhood of $o_F$. Since $f$ is continuous at $o_E$, there exists a neighbourhood $U \subseteq E$ of $o_E$ with $f(U) \subseteq f(V)$. Then $U + e$ is a neighbourhood of $e$, and for all $x \in U + e$, by writing $x = u + e$ where $u \in U$ we have that:

\begin{align} \quad f(x) = f(u + e) = f(u) + f(e) \in f(U) + f(e) \subseteq V + f(e) \end{align}

- So $f(U + e) \subseteq V + f(e)$. So $f$ is continuous at $e$, and is thus $f$ is continuous on $E$. $\blacksquare$

Proposition 3: Let $E$ and $F$ be normed spaces over $\mathbf{F}$. Then a linear operator $f : E \to F$ is continuous if and only if there exists a constant $M > 0$ such that $\| f(x) \| \leq M \| x \|$ for all $x \in E$. |

**Proof:**$\Rightarrow$ Suppose that $f : E \to F$ is continuous. Then $f$ is continuous at $o_E$. So for the neighbourhood $\{ y : \| y \| \leq 1 \} \subseteq F$ of $f(o_E) = o_F$, there exists a neighbourhood $U$ of $o_E$ such that $f(U) \subseteq \{ y : \| y \| \leq 1 \}$. But since $U$ is a neighbourhood of $o_E$, there exists a $\delta > 0$ such that $\{ x : \| x \| \leq \delta \} \subseteq U$. Thus if $\| x \| \leq \delta$ then $\| f(x) \| \leq 1$.

- So for each $x \in E$ with $x \neq 0$, we have that $\frac{\delta x}{\| x \|}$ is such that $\left \| \frac{\delta x}{\| x \|} \right \| = \delta$, so :

\begin{align} \quad \left \| f \left ( \frac{\delta x}{\| x \|} \right ) \right \| \leq 1 \end{align}

- Which implies that:

\begin{align} \quad \| f(x) \| \leq \frac{1}{\delta} \| x \| \end{align}

- Trivially the inequality above also holds for $x = 0$. So, set $M := \frac{1}{\delta}$. Then $\| f(x) \| \leq M \| x \|$ for all $x \in E$.

- $\Leftarrow$ Suppose that there exists an $M > 0$ such that $\| f(x) \| \leq M \| x \|$ for all $x \in E$. Let $V$ be a neighbourhood of $f(o_E) = o_F$. Then there exists an $\epsilon > 0$ such that $\{ y \in F : \| y \| \leq \epsilon \} \subseteq V$.

- Let $U = \{ x : \| x \| \leq \frac{\epsilon}{M} \}$. Then $U$ is a neighbourhood of $o_E$, and for all $x \in U$ we have that:

\begin{align} \quad \| f(x) \| \leq M \| x \| \leq M \cdot \frac{\epsilon}{M} = \epsilon \end{align}

- and so $f(x) \in \{ y : \| y \| \leq \epsilon \} \subseteq V$. So $f(U) \subseteq V$. Thus $f$ is continuous at the origin. By the previous proposition, this implies that $f$ is continuous. $\blacksquare$

## Continuity of a Set of Linear Operators

Definition: Let $E$ and $F$ be topological vector spaces and let $\{ f_i : i \in I \}$ be a collection of linear operators from $E$ to $F$. Then $\{ f_i : i \in I \}$ is Continuous if each $f_i$ is continuous. |

Note that the collection of linear operators $\{ f_i : i \in I \}$ is continuous if for every neighbourhood $V \subseteq F$ of $o_F$ we have that for each $i \in I$ there is a neighbourhood $U_i \subseteq E$ of the $o_E$ such that $f_i(U_i) \subseteq V$.

## Equicontinuity of a Set of Linear Operators

Definition: Let $E$ and $F$ be topological vector spaces and let $\{ f_i : i \in I \}$ be a collection of linear operators from $E$ to $F$. Then $\{ f_i : i \in I \}$ is Equicontinuous if for every neighbourhood $V \subseteq F$ of $o_F$ there is a neighbourhood $U \subseteq E$ of $o_E$ such that $f_i(U) \subseteq V$ for each $i \in I$. |