----- Let $V,W$ be [[norm|normed]] [[vector space|vector]] [[normed vector space|spaces]] over a common base field $\mathbb{F}$ (either $\mathbb{R}$ or $\mathbb{C}$). Let $T:V\to W$ be a [[linear map|linear map]]. > [!proposition] Proposition. ([[characterizing continuity of linear maps]]) > The following are equivalent: > 1. $T$ is [[operator norm|bounded]]; > 2. $T$ is [[continuous]]; > 3. $T$ is [[continuous]] at the origin; > 4. There exists $f \in V$ such that $T$ is [[continuous]] *[[metric-space continuity|at]]* $f$; > 5. $T$ is [[uniformly continuous]]; indeed, [[Lipschitz continuous|Lipschitz]][^1] > 6. $T ^{-1}\big( B_{r}(0) \big)$ is an open subset of $V$ for some $r > 0$. > > Furthermore: if $V$ is [[dimension|finite-dimensional]], $T$ is [[continuous]]. If not, $T$ need not be [[continuous]] — but constructing examples appears to require the [[Axiom of Choice]] (in the form of [[Zorn's lemma]] to prove [[every vector space has a basis|Hamel bases always exist]]) once we assume $V$ is [[complete]]. ^proposition **A discontinuous linear map ('trivial' — non-Banach).** In $\ell^{2}(\mathbb{N})$ consider $X=\span \{ e_{n}: n \in \mathbb{N} \}$ and $Y=\mathbb{\mathbb{F}}$, $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. Then define $Te_{n}=n$ and linearly extend. Clearly $T$ is unbounded. This example is 'trivial' because although $X$ is [[dense]] in $\ell^{2}$[^2] it is not [[complete]]/[[closed set|closed]].[^3] **A discontinuous linear map (Banach).** Let $X,Y$ be be normed vector spaces and assume $\text{dim }X=\infty$ and $Y \neq \{ 0 \}$. Then there exists an unbounded linear operator $T:X \to Y$. Indeed, let $A=\{ x_{n}:n \in \mathbb{N} \}$ be a [[linearly independent]] subset of $X$. By [[every vector space has a basis]] there exists a [[basis|Hamel basis]] $B$ for $X$ such that $A \subset B$. Define the functional $f$ by $f(x_{n})=n \|x_{n}\|$, $n \geq 1$, and $f(x)=0$ for all $x \in B - A$ and extend linearly. Then $f$ is unbounded. [^2]: To see this, it suffices by [[neighborhood-basis characterization of set closure]] to witness that any ball of radius $\varepsilon>0$ about and $(x_{1},x_{2},\dots) \in X$ nontrivially intersects $X$, in other words, that there exists a finite sequence arbitrarily close to any sequence in $\ell^{2}$. This is true essentially by definition of [[series]]: for $N \in \mathbb{N}$ define $x^{(N)}:=x_{1}e_{1}+\dots+x_{N}e_{N} \in X$. Then since $(x_{k}) \in \ell^{2}$, $\sum_{k=1}^{\infty} |x_{k}|^{2}$ converges, from which it follows that there exists $N \in \mathbb{N}$ such that $\sum_{k=N}^{\infty} |x_{k}|^{2}<\varepsilon^{2}$, whence $\|(x_{k})_{k=1}^{\infty}- x_{k}^{(N)}\|_{2}= \sqrt{ \sum_{k=N}^{\infty} |x_{k}|^{2} } < \varepsilon.$Thus $\overline{X}=\ell^{2}$. [^3]: One can see $X$ is not closed e.g. by appealing to [[the sequence lemma]]: the element $(x_{k})=\left( \frac{1}{k} \right)_{k=1}^{\infty} \in \ell^{2}$ is not in $X$, but it is converged to by finite-support vectors (elements of $X$) $(1,0,\dots), \left( 1, \frac{1}{2}, 0, \dots \right), (1, \frac{1}{2}, \frac{1}{3}, 0, \dots),\dots$ of elements in $X$: if $x^{(N)}=\left( 1, \frac{1}{2}, \frac{1}{3}, \dots ,\frac{1}{N},0,\dots \right)$, then $\|(x_{k})-x^{(N)}\|_{2}= \sqrt{ \sum_{k=N+1}^{\infty} \frac{1}{k^{2}} } \xrightarrow{N \to \infty}0.$ [^1]: In this case, the [[Lipschitz continuous|Lipschitz constant]] is precisely $\|T\|$. **$(2) \implies (3)$.** Obvious. **$(3) \implies (4).$** Obvious. **$(1) \iff (5).$** Suppose $\|T\|<\infty$. Then for all $f,g \in V$, $\|Tf-Tg\|=\|T(f-g)\| \leq \|T\| \|f-g\|,$showing that $T$ is [[Lipschitz continuous]] with constant $L=\|T\|$. The converse is immediate. Continuity at $0$ gives the existence of $\delta>0$ such that $\|g\|<\delta \implies \|Tg\|<1$ for $g \in V$. For any $g \neq 0$, $\|Tg\|= \frac{\|g\|}{\delta} \|T\left( \delta \frac{g}{\|g\|} \right)\| \leq \frac{1}{\delta} \|g\|,$ hence $\|T\|\leq \frac{1}{\delta} < \infty$. > [!proof]- Proof. ([[characterizing continuity of linear maps]]) > > **$(1) \iff (2)$.** > Suppose $V,W$ are [[normed vector space|normed vector spaces]] and $T \in \text{Hom}(V,W)$. Contrapositively suppose $T$ is not bounded. We'll construct a sequence tending to zero but whose image does not (recall that linear maps send 0 to 0), implying noncontinuity. Indeed, there exists a [[sequence]] $f_{1},f_{2},\dots$ in $V$ such that $\|f_{k}\| \leq 1$ for all $k \in \mathbb{N}$ and $\lim_{k \to \infty}\|Tf_{k}\|=\infty$.[^1] Then $\lim_{k \to \infty} \frac{f_{k}}{\|Tf_{k}\|}=0,$ but $\ T\left( \frac{f_{k}}{\|Tf_{k}\|} \right) \not \to 0$ where the nonconvergence holds because each term has unit norm. > Conversely, suppose $T$ is bounded and $(f_{k}) \to f$ is a [[sequence]] in $V$. Then $T\big( (f_{k}) \big) \to Tf$, as follows from the fact $\|Tf_{k}-Tf\| = \|T(f_{k}-f)\| \leq \|T\| \|f_{k}-f\|.$ **"Furthermore..."** Let $T: V\to W$ be a [[linear map|linear map]] between normed vector spaces with $V$ [[dimension|finite-dimensional]], $\text{dim } V=n$. Fixing a [[basis]] $\{ e_{i} \}$ of $V$, we have for $v \in V$ (coordinates: $(v_{i})$) the bound (for any norm) $\begin{align} \|Tv\| &= \|\sum_{i=1}^{n} v_{i} Te_{i}\| \\ & \leq \sum_{i=1}^{n} |v_{i}| \|Te_{i}\| \\ & \leq \underbrace{ (\max \|Te_{i}\| ) }_{ := c } \left( \sum_{i=1}^{n} | v_{i}| \right) \end{align}$ For the [[sup norm]] in particular we further obtain $\|Tv\|_{\infty} \leq c n \|v\|_{\infty}.$ This bound implies continuity at zero in the sup norm: fixing $\varepsilon>0$ and defining $\delta_{0}:=\frac{\varepsilon}{cn }$, one has for $\|v\|_{\infty}<\delta_{0}$ that $\|Tv\|_{\infty} \leq cn \|v\|_{\infty} < cn \delta_{0} =\varepsilon.$ Since all [[norm|all norms are equivalent on a finite-dimensional vector space]], and a [[linear map|linear map]] is [[continuous]] iff it is [[continuous]] [[characterizing continuity of linear maps|at zero]], we are done. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```