> [!definition] > > Let $\cx$ and $\cy$ be [[Normed Vector Space|normed spaces]], then $L(\cx, \cy)$ is the space of all [[Bounded Linear Map|bounded linear maps]] from $\cx$ to $\cy$. $L(\cx, \cy)$ with the **operator norm** > $ > \begin{align*} > \norm{T} &= \sup_{\norm{x} = 1}\norm{T(x)} \\ > &= \sup\bracs{\frac{\norm{T(x)}}{\norm{x}}: x \ne 0} \\ > &= \inf\bracs{C: \norm{T(x)} \le C\norm{x} \ \forall x \in \cx} > \end{align*} > $ > forms a normed space. > > *Proof*. Let $T \in L(\cx, \cy)$, > $ > \begin{align*} > s &= \sup\bracs{\frac{\norm{T(x)}}{\norm{x}}: x \ne 0} \\ > c &= \inf\bracs{C: \norm{T(x)} \le C\norm{x} \ \forall x \in \cx} > \end{align*} > $ > Then for any $\varepsilon > 0$ there exists $x \in \cx$ such that $\frac{T(x)}{\norm{x}} + \varepsilon > s$, where $\norm{T(x)} > (s - \varepsilon)\norm{x}$. Therefore $c \ge s - \varepsilon$ for all $\varepsilon > 0$, and $c \ge s$. > > On the other hand, for all $\varepsilon > 0$ there exists $x \in \cx$ such that $\norm{T(x)} > (c - \varepsilon)\norm{x}$, where $\frac{\norm{T(x)}}{\norm{x}} > c - \varepsilon$. Therefore $s \ge c - \varepsilon$ for all $\varepsilon > 0$ and $s \ge c$. > > Let $\lambda \in \complex$, then > $ > \bracs{\frac{\norm{\lambda T(x)}}{\norm{x}}: x \ne 0} = \abs{\lambda}\bracs{\frac{\norm{T(x)}}{\norm{x}}: x \ne 0} > $ > so $\norm{\lambda T} = \abs{\lambda}\norm{t}$. Now let $S, T \in L(\cx, \cy)$, then > $ > \begin{align*} > \frac{\norm{(S + T)(x)}}{\norm{x}} \le \frac{\norm{S(x)}}{\norm{x}} + \frac{\norm{T(x)}}{\norm{x}} \quad \forall x \in \cx > \end{align*} > $ > Since this holds for any $x$, $\norm{S + T} \le \norm{S} + \norm{T}$. Therefore $\norm{\cdot}$ is a norm. > [!theorem] > > Let $\cx$ be a [[Banach Space|Banach space]], then the space $L(\cx, \cx)$, with the product being composition, is a [[Banach Algebra|Banach algebra]]. > [!theorem] > > Let $\cx$ and $\cy$ be normed spaces and $S, T \in L(\cx, \cy)$, then > $ > \norm{S(x) - T(x)} \le \norm{S - T}\cdot \norm{x} \quad \forall x \in \cx > $ > *Proof*. Let $x \in \cx$ such that $\norm{x} = 1$, then > $ > \norm{S(x) - T(x)} = \norm{(S - T)(x)} \le \norm{S - T} > $ > For any $x \ne 0$, > $ > \begin{align*} > \norm{S(x) - T(x)} &= \norm{(S - T)(x)} \\ > &= \norm{x} \cdot \norm{(S - T)\paren{\frac{x}{\norm{x}}}} \\ > &= \norm{x} \cdot \norm{S - T} > \end{align*} > $ > [!theorem] > > Let $\cx$ and $\cy$ be normed spaces. If $\cy$ is a [[Banach Space]], then so is $L(\cx, \cy)$. > > *Proof*. > > ### Construction > > Let $\seq{T_n}$ be a [[Cauchy Sequence|Cauchy sequence]] in $L(\cx, \cy)$, then for any $\varepsilon > 0$, there exists $N \in \nat$ such that $\norm{T_m - T_n} < \varepsilon$ for all $m, n \ge N$. Let $x \in \cx$, then > $ > \norm{T_m(x) - T_n(x)} \le \norm{T_m - T_n} \norm{x} = \varepsilon \norm{x} > $ > so $\seq{T_n(x)}$ is also Cauchy. As $\cy$ is a Banach space, $T_n(x)$ [[Limit|converges]]. > > > ### Linearity > > Let $T: \cx \to \cy$ with $x \mapsto \limv{n}T_n(x)$. Let $x, y \in \cx$, and $\lambda \in \complex$, then since addition and scalar multiplication are continuous, > $ > \begin{align*} > \limv{n}T_n(\lambda x + y) &= \limv{n}\braks{\lambda T_n(x) + T_n(y)} \\ > &= \lambda \limv{n}T_n(x) + \limv{n}T_n(y) > \end{align*} > $ > we have that $T$ is linear. > > > ### Boundedness > > Now, as $\seq{T_n}$ is Cauchy, there exists $N \in \nat$ such that $\norm{T_m - T_n} \le 1$ for all $m, n \ge N$. Let $x \in \cx$, take $C_N$ such that $\norm{T_N(x)} \le C_N\norm{x}$ for all $x \in \cx$, then > $ > \begin{align*} > \norm{T_n(x)} &\le \norm{T_N(x)} + \norm{T_n(x) - T_N(x)} \\ > &\le \norm{T_N(x)} + \norm{T_n - T_N} \cdot \norm{x} \\ > &\le C_N\norm{x} + \norm{x} \\ > &= (C_N + 1)\norm{x} > \end{align*} > $ > and $\norm{T_n(x)} \le (C_N + 1) \norm{x}$ for all $x \in \cx$ and $n \in \nat$. Therefore > $ > \norm{T(x)} = \norm{\limv{n}T_n(x)} = \limv{n}\norm{T_n(x)} \le (C_N + 1)\norm{x} > $ > since the norm is continuous with respect to itself. > > > ### Convergence > > Lastly, let $\varepsilon > 0$, then there exists $N \in \nat$ such that > $ > \norm{T_m - T_n} < \varepsilon \quad \forall m, n \ge N > $ > Now, let $n \ge N$ and $x \in \cx$, then > $ > \norm{T_N(x) - T_n(x)} \le \norm{T_N - T_n}\norm{x} \le \varepsilon\norm{x} > $ > meaning that > $ > \norm{T(x) - T_N(x)} = \limv{n}\norm{T_n(x) - T_N(x)} \le \varepsilon \norm{x} > $ > and we have $\norm{T - T_n} \le \varepsilon$. Therefore $\norm{T - T_n} \to 0$. > [!theorem] > > Let $\cx$ and $\cy$ be normed spaces, then $L(\cx, \cy)$ is an [[Algebra over Ring|algebra]] with composition being the product. If $L(\cx, \cy)$ is [[Complete Metric Space|complete]], then it is a [[Banach Algebra|Banach algebra]]. > > *Proof*. Let $x \in \cx$ and $R, S, T \in L(\cx, \cy)$, then > $ > R(aS + T)x = R(aSx + Tx) = aRSx + RTx > $ > and > $ > (aR + S)T(x) = aRTx + STx > $ > so the composition is [[Multilinear Map|bilinear]]. Now, > $ > \norm{STx} \le \norm{S}\norm{Tx} \le \norm{S} \norm{T} \norm{x} > $ > so $ST \in L(\cx, \cy)$ and $\norm{ST} \le \norm{S} \norm{T}$. > [!theorem] > > Let $\cx, \cy$ be normed spaces, then the map $(T, x) \mapsto Tx$ is continuous from $L(\cx, \cy)$ to $\cy$. > > *Proof*. Let $T \in L(\cx, \cy)$, $x \in \cx$, and $\varepsilon > 0$, then there exists > $ > \delta \in (0, 1): \delta(\norm{T} + \norm{x} + 1) < \varepsilon > $ > such that whenever $\norm{(T, x) - (T', x')} < \delta$, we have > $ > \begin{align*} > \norm{Tx - T'x'} &= \norm{Tx - Tx' + Tx' - T'x'} > \\ > &\le \norm{Tx - Tx'} + \norm{Tx' - T'x'} \\ > &\le \norm{T}\norm{x - x'} + \norm{T - T'}\norm{x'} > \\ > &\le \delta(\norm{T} + \norm{x} + \delta) \\ > &< \varepsilon > \end{align*} > $