> [!definition] > > Let $\cx$ and $\cy$ be [[Normed Vector Space|normed spaces]]. A [[Linear Transformation|linear map]] $T: \cx \to \cy$ is **bounded** if there exists $C > 0$ such that > $ > \norm{T(x)} \le C\norm{x} \quad \forall x \in X > $ > [!theorem] > > Let $\cx$ and $\cy$ be normed spaces and $T: \cx \to \cy$ be a linear map. The following are equivalent: > - $T$ is [[Continuity|continuous]]. > - $T$ is continuous at $0$. > - $T$ is bounded. > > *Proof*. ($b \Rightarrow a$) Suppose that $T$ is continuous at $0$, then > $ > \forall \varepsilon > 0, \exists \delta > 0: \norm{T(x)} < \varepsilon \quad \forall x \in B(0, \delta) > $ > For any $x \in \cx$ choose $y$ such that $\norm{x - y} < \delta$, then > $ > \norm{T(x) - T(y)} = \norm{T(x - y)} < \delta > $ > and $T$ is continuous at $x$. > > ($b \Rightarrow c$) Suppose that $T$ is continuous at $0$, and choose $\delta > 0$ such that $\norm{T(x)} \le 1$ whenever $\norm{x} \le \delta$. > > Let $x \in \cx$, $x \ne 0$, then since $\norm{\frac{\delta x}{\norm{x}}} \le \delta$, > $ > \begin{align*} > \norm{T(x)} &= \norm{T\paren{\frac{x\norm{x}}{\norm{x}}}} \\ > &= \norm{x} \cdot \norm{T\paren{\frac{\delta x}{\delta\norm{x}}}} \\ > &= \frac{\norm{x}}{\delta} \cdot \norm{T\paren{\frac{\delta x}{\norm{x}}}} \\ > &\le \frac{\norm{x}}{\delta} > \end{align*} > $ > and $T$ is bounded. > > ($c \Rightarrow b$) Suppose that $T$ is bounded with $\norm{T(x)} \le C\norm{x}$, then for any $\varepsilon > 0$, there exists $\delta = \frac{\varepsilon}{C}$, and > $ > \norm{T(x)} \le C\norm{x} < \frac{C\varepsilon}{C} = \varepsilon \quad \forall x \in B(0, \delta) > $ > so $T$ is continuous at $0$. > [!theorem] > > ![[bound-linear.png|300]] > > Let $\cx, \cy$ be normed spaces and $T: \cx \to \cy$ be a linear map. If $T$ is bounded on an [[Open Set|open set]] $U$, then $T$ is bounded. > > *Proof*. Let $U$ be an open set such that $\norm{Tx} \le M$ for all $x \in U$. Let $x \in U$, then there exists $\varepsilon > 0$ such that $\ol{B(x, \varepsilon)} \subset U$. We translate the ball back to the origin by removing $x$ from the coordinate and maintaining the bound. > > Let $y \in \cx$ such that $\norm{y} = 1$, then $x + \varepsilon y \in \ol{B(x, \varepsilon)}$ and > $ > \begin{align*} > \varepsilon\norm{Ty} = \norm{T\varepsilon y} &\le \norm{T(x + \varepsilon y)} + \norm{x} \le 2M \\ > \norm{Ty} &\le \frac{2M}{\varepsilon} > \end{align*} > $ > Therefore $T$ is bounded. > [!definition] > > Let $\cx, \cy$ be normed spaces and $T \in L(\cx, \cy)$ be a bounded linear map. $T$ is **invertible** (is an [[Isomorphism|isomorphism]]) if it is bijective and $T^{-1}$ is bounded (continuous).