Let $V$ be a $K$–vector space, let $V^*$ be its (algebraic) dual space, and let $L(V, V; K)$ denote the space of bilinear maps from $V \times V \to K$.
Does $V^* \otimes V^* \cong L(V, V; K)$ hold when $V$ is infinite dimensional? If so, please give a proof using the universal property.
[By $\cong$, I mean not only vector space isomorphism, but isomorphism as tensor product spaces over $V^* \times V^*$. The "natural" multilinear map from $V^* \times V^* \to L(V, V; K)$ is the one taking $(\omega_1, \omega_2) \mapsto \omega_1(\cdot)\omega_2(\cdot)$. I'm asking whether $L(V, V; K)$ along with this multilinear map is a tensor product space over $V^* \times V^*$, in the sense of satisfying the universal property.]
If $V$ is finite dimensional, then this isomorphism and several related isomorphisms hold. I know that in general, these isomorphisms rely on $V \simeq V^*$, which only holds when $V$ is finite dimensional. However, the "unary" version of the above identity, $V^* \simeq L(V; K)$ holds (by definition) regardless of the dimension of $V$, and if the "binary" version above holds, it would be a nice way to see why $V \otimes V \simeq L(V^*, V^*; K)$ also holds in (and only in) the finite dimensional case (using $V \simeq V^{**})$.
My approach was to consider the map $\phi: V^* \times V^* \to L(V, V; K)$ defined by $\phi(\omega_1,\omega_2)(v_1, v_2) = \omega_1(v_1) \omega_2(v_2)$. Then $\phi$ is multilinear, so the universal property gives a linear map $\tilde\phi : V^* \otimes V^* \to L(V, V; K)$ such that $\phi = \tilde\phi \circ \otimes$. If we can show $\tilde\phi$ is an isomorphism, we are done. But I can't show this, and don't even know if it is true.
Your map may fail to be surjective. So, I suspect the isomorphism does not hold.
Let $V$ denote the space of real sequences with at most finitely many elements. Let $\alpha \in L(V,V; K)$ denote the map $$ \alpha[(x_1,x_2,\dots),(y_1,y_2,\dots)] = \sum_{k=1}^\infty x_ky_k. $$ I claim that this cannot be expressed as a finite sum of the form $\alpha = \sum_{k=1}^n \omega_1^{(k)} \otimes \omega_2^{(k)}$.
Suppose for the purpose of contradiction that such a decomposition of $\alpha$ exists. We can assume without loss of generality that the $\omega_i^{(k)}$ are linearly independent (and thus have distinct kernels). The functionals $\omega_1^{(k)}$ each have one-dimensional cokernels. It follows that the quotient space $$ V/(\ker(\omega_1^{(1)}) \cap \cdots \cap \ker(\omega_1^{(n)})) \cong V/\ker(\omega_1^{(1)}) \oplus \cdots \oplus V/\ker(\omega_1^{(n)}) $$ is also finite dimensional. Thus, $\ker(\omega_1^{(1)}) \cap \cdots \cap \ker(\omega_1^{(n)})$ must be infinite dimensional. Select any non-zero $x \in \ker(\omega_1^{(1)}) \cap \cdots \cap \ker(\omega_1^{(n)})$. From the decomposition of $\alpha$, we find that $\alpha(x,y) = 0$ holds for all $y$ in $V$. However, this clearly does not hold for our $\alpha$ as it was originally defined; for instance, we can see that $\alpha(x,x) \neq 0$ will hold for all non-zero $x$ (for other fields, take $y$ equal to the standard basis elements of $V$). Thus, we have reached a contradiction.