Are there even discontinuous linear functionals?

For me it was hard to think of an example of a discontinuous linear function. Certainly in the finite dimensional case all linear functions are continuous, because they are - in a sense - simply matrix multiplications.

Therefore, we have to think big - infinite dimensional big. Consider all bounded infinitely differentiable functions on \(\R\) equipped with the supremums norm \(\norm{.}_\infty\). Define the sequence \[f_n = x \mapsto \frac{1}{n}\sin(n x).\] It is easy to see that \(\lim_{n \to \infty} f_n = 0\).

But for the derivatives \(f_n' = x \mapsto \cos(n x)\) the limit does not exist. Therefore, the linear operator \(\frac{d}{dx}\) is not continuous.

How can we characterise continuous linear functions?

To be able to speak of continuity we at least need a topology.

A real (complex) vector space \(X\) with Hausdorff topology \(\T\) is said to be a topological vector space if addition and scalar multiplication is continuous with respect to \(\T\).

The dual space \(X'\) is the set of all continuous linear functions \(f:X \to \C\). It is a subspace of all functionals \(X^*\).

Just like all functionals on \(\R^n\) are continuous in the case of finite dimensional topological vector spaces this also holds.

Theorem.

Let \(\left< X, \T \right>\) be a \(n\) dimensional topological vector space then \(X' = X^*\).

Proof.

Consider the normed space \(\left<\C^n, \norm{.}_2 \right>\). Let \(\{e_i : i=1,\dots,n\}\) be the standard basis of \(\C^n\) and let \(f \in \C^{n*}\). We have \[|f(z)| = |f(\sum_{i=1}^n \alpha_i e_i)| = \sum_{i=1}^n |\alpha_i ||f(e_i)|\le C \norm{z}_2.\] Therefore, \({\C^{n}}' = \C^{n*}.\)

Now it is sufficient to show that \(X\) is homeomorphic to \(\C^{n}\).

Let \(\psi: \C^n \to X\) be an isomorphism. Again \[\psi(z) = \sum_{i=1}^n \alpha_i \psi(e_i).\] As \(|\alpha_i| \le \norm{z}_2\) the coefficients depend continuously on \(z\). As addition and scalar multplication on \(X\) are continuous \(\psi\) is continuous.

Now we have to show that \(\psi^{-1}\) is continuous. As it is linear it is enough to show continuity at \(0\).

Let \(S^{n-1} = \{z \in \C^n : \norm{z}_2 = 1 \}\) which is compact. Thus \(K:= \psi(S^{n-1}) \subseteq X\) is also compact and since T2 closed. As \(\operatorname{ker}(\psi) = 0\) we have \(0 \in X \setminus K\). Choose a circled open neigbourhood of \(0\), \(V \subseteq X \setminus K\).

Let \(E := \psi^{-1}(V)\). This set is circled and \[E \cap S^{n-1} = \psi^{-1}(V \cap K) = \emptyset.\] Thus \(E \subseteq U_1^{\C^n}(0)\).

Let \(\epsilon > 0\). \[\psi^{-1}(\epsilon V) = \epsilon \psi^{-1}(V) = \epsilon E \subseteq U_\epsilon^{\C^n}(0)\] which proves continuity at \(0\).

\(\Box\)

In the introduction we found an example for a discontinuous function in an infinite dimensional vector space. Interestingly enough, this is always possible.

Proposition.

Let \(\left< X, \norm{.} \right>\) be a infinite dimensional normed vector space then \(X^* \setminus X' \neq \emptyset\).

Proof.

Let \(\{b_n : n \in \mathbb{N}\}\) be a linearly independent set.

By the theorem of linear continuation let \(f\) be the linear function with \[f(b_n) = n \cdot \norm{b_n}.\] This map is unbounded and thus discontinuous.

\(\Box\)

Note that for any vector space \(X\) we can find a topology such that all functionals are continuous. In particular let \(\T = \sigma(X,X^*)\) be the weak topology with respect to \(X^*\). Then \(\left< X, \T \right>\) is a locally convex topological vector space where every functional is continuous.

There even exist a topological vector space where there is no continuous functional.

Proposition.

Let \(p \in (0,1) \) and let \[\Delta f = \int_{(0,1)} |f(t)|^p d\lambda(t).\] Then \(L^p(0,1) = \{f:(0,1)\to \C : \Delta f < \infty\}\) is a vector space with metric \[d(f,g) = \Delta(g-f).\] It holds that \({L^p(0,1)}' = \{0\}\).

Proof.

Claim.

If a \(V\), a neighbourhood of \(0\), is convex, then \(V = L^p(0,1)\).

Proof.

Let \(f\in L^p(0,1)\) and \(U_r^\Delta(0) \subseteq V\). Choose \(n \in \mathbb{N}\) such that \(n^{p-1} \Delta f < r\).

The measure \[\mu(A) = \int_A |f(t)|^p d\lambda(t)\] is finite and thus \[x \mapsto \mu((0,x))\] is continuous. Therefore, we can choose \(x_j\) such that \(\mu(0,x_j) = \frac{j}{n}\Delta f\).

Define \(g_i = n f I_{(x_{i-1}, x_j]}\). It holds that \(f = \frac{1}{n}\sum_{i=1}^n g_i\) and \[\Delta g_i = \int_{(x_{i-1}, x_j]} |nf(t)|^p d\lambda(t) = n^{p-1} \Delta f < r.\] Thus \(g_i \in V\) and as a convex combination \(f \in V\).

\(\Box\)

Suppose there is a \(\psi \in L^p(0,1)'\) such that there is a \(f \in L^p(0,1)\) with \(\psi(f) = c \neq 0\).

Choose \(\epsilon = \frac{|c|}{2}\). Now for any \(\delta > 0 \) it holds that \[\psi(U_\delta^\Delta(0)) = \psi(L^p(0,1)) \ni c.\] So we have \[\psi(U_\delta^\Delta(0)) \not\subseteq U_\epsilon^\Delta(0)\] which makes \(\psi\) discontinuous.

\(\Box\)