Why is this inclusion of dual of Banach spaces wrong? Ive been struggling the last days on this paradox, please I need help! Let
$$E\subset F$$
be two Banach spaces equipped with the same norm. Some people told me that
$$F^* \subset E^*$$
with $E^*$ denoting the dual space of $E$ (space of continuous linear functional defined on $E$). But this statement is wrong. (Why? Let $t_i:F^*\to E^*$ be defined as the restriction of the functionals of $F^*$ on $E$, this $t_i$ is surjective so $E^*$ is smaller than $F^*$ thus the inclusion cannot hold).
However, their proof for the inclusion $F^*\subset E^*$ is : let $T\in F^*$, $T$ is defined on $F$ thus on $E$, it is linear and continuous on $E$, therefore, $T\in E^*$.
1) What is wrong in their proof? It seems so true!
2) Is $E^*\subset F^*$? And then what the inclusion "$\subset$" would mean? (because if $T\in E^*$, it is not direct to make it belong to $F^*$!)
Thanks for your help.
 A: The restriction homomorphism $\varrho \colon F^\ast \to E^\ast$ - which is the adjoint of the inclusion $\iota \colon E\hookrightarrow F$ - is in general not injective. It is injective if and only if $E$ is dense in $F$, the kernel is the annihilator of $E$,
$$\ker\varrho = E^\perp = \{ \lambda \in F^\ast : E\subset\ker\lambda\}.$$
By the Hahn-Banach theorem(s), $\varrho$ is surjective, and hence it induces an isometric isomorphism
$$\overline{\varrho} \colon F^\ast/E^\perp \xrightarrow{\cong} E^\ast.$$
In general, you have no natural inclusion $E^\ast \hookrightarrow F^\ast$ either. You have such an inclusion if $\overline{E}$ is a complemented subspace of $F$, i.e. there is a closed subspace $G\subset F$ with $F \cong \overline{E}\oplus G$, then you have a decomposition $F^\ast \cong E^\ast \times G^\ast$, and that gives you a natural inclusion of $E^\ast$ into $F^\ast$.
A situation where you have "$F^\ast \subset E^\ast$", or, more precisely, a natural continuous injection $F^\ast \hookrightarrow E^\ast$, is when $E$ is a dense subspace of $F$ (so the restriction is injective), but the topology on $E$ is (strictly) finer than the topology induced by $F$. A typical example of such a situation is $H^1(\mathbb{R}^n) \subset L^2(\mathbb{R}^n)$, where $H^1(\mathbb{R}^n)$ is the Sobolev space of square integrable functions whose weak first-order derivatives are also square integrable. Then the linear forms
$$\lambda_g \colon f \mapsto \int_{\mathbb{R}^n} \frac{\partial f}{\partial x_i}(x)\overline{g(x)}\,dx$$
for $g \in L^2(\mathbb{R}^n)$ belong to $H^1(\mathbb{R}^n)^\ast$, but unless $g = 0$, are not continuous in the topology induced by $L^2$, so $\iota^\ast(F^\ast) \subsetneqq E^\ast$ in this situation.
