Are there densely defined symmetric operators from a Banach space into its algebraic dual which do not map into the continuous dual?

Question

The notion of symmetric operator on Hilbert space can be generalized to locally convex spaces as follows: A map $T \colon D \to X^a$ between a subspace $D$ of a locally convex topological vector space $X$ and its algebraic dual $X^a$ is called symmetric, if

$$(Tx,y)_{X^a,X} = (x,Ty)_{X,X^a}$$,

where the brackets denote duality pairings. This condition automatically ensures that $T$ is linear.

For the special case where $X$ is a Banach space $B$ and $T$ is defined a) on a dense linear subspace or b) on all of $B$, does this symmetry condition ensure that $T$ must map into the continuous dual $B^{\ast}$?

As the trivial example of an operator defined on $\{0\}$ shows, this is not true in general.

Here are some related results:

1. If $T \colon B \to B^{\ast}$ is symmetric, then it is continuous (this is a generalization of the Helling-Toeplitz theorem on Hilbert spaces and can be found as Proposition 1.1 on p. 145 in the book "Probability on Banach spaces" by Vakhania, Tarieladze and Chobanyan). From this

2. If $T \colon B^{\ast} \to B^{\ast a}$ is the covariance operator of some Borel measure on $B$ (so in particular positive symmetric), then $T$ maps into $B^{\ast\ast}$. This is claimed (referring to an article in Russian and not easily available) on page 12, before stating Theorem 6 of Covariance operators of probability measures in locally convex spaces by Vakhania and Tarieladze.

Answer 1

Let $H$ be any infinite-dimensional Hilbert space, and let $\psi:H\to\mathbb C$ be an unbounded linear functional (these exist in ZFC). Define $T:H\to H^a$ by $$ Tx=\psi(x)\,\psi. $$ Then $T$ is symmetric, defined everywhere, and $TH\subset H^a\setminus H^*$.

If you don't want Choice, then you lose a lot of functional analysis; but in any case the above can be done on a dense subspace.