I have the following: Suppose that $E $ is an infinite dimensional vector space. Then that there exists a dual space $E^*$ such that the natural injection $\varphi : E^* \rightarrow L (E)$ given by $\varphi (x^*) = \langle - , x^* \rangle : E \rightarrow \Gamma$ is not surjective, where $L (E) $ is the set of linear mappings $\varphi : E \rightarrow \Gamma $ $($ $\langle -,- \rangle$ takes values in $\Gamma$ $)$
(In my context all vector spaces are defined over a fixed, but arbitrarily chosen field $ \Gamma $ of characteristic $ 0 $)
Is this statement true? I have been looking for a counterexample because I suspect that it is not true since assuming surjectivity for the function, there would be an isomorphism between the two dual spaces and then their dimensions are equal. Could you get me out of the mistake if I'm wrong.
Definition: If $E, E^*$ is a pair of vector space and if a fixed non-degenerate bilinear function, $\langle , \rangle$, in $E^* \times E$ is defined. Then $E$ and $E^*$ will be called dual with respect to the bilinear function $\langle , \rangle $.
I don't know how to use the fact that $ E $ is an infinite dimensional vector space. Could you give me any suggestion please?
A very important theorem in linear algebra that is rarely taught is:
Theorem: A vector space has the same dimension as its dual if and only if it is finite dimensional.
Edit: At this point I am not sure, if $M_{log}$ is continous w.r.t. $\lVert \cdot \rVert_1$. I expect $M_{log}(C^0([1,\infty))\subset L^1([1,\infty))$, since I expect $f\in C^0([1,\infty)) \cap L^1([1,\infty))$ to vanish at least of power $x^{1+c}$ for some $c$ as $x\rightarrow\infty$ (which was my initial thought), yet not uniformly.
I don't delete this post to keep this edit as a red flag.
Let $E:=L^1([1,\infty))$ be a real vector space and hence $E^*=L^\infty([1,\infty))$. In this setting the natural injection $\phi$ maps an element $g\in E^*$ to the multiplication operator $M_g$, which is of course an element of $L(E)$. Yet $M_{log}$, i.e. the multiplication with the logarithm is an element of $L(E)$ which is given by an unbounded function and hence $M_{log}\not\in \phi(E^*)$.