Suppose that $X$ is a finite-dimensional Banach space. Let $T\in L(X)$ be an operator on $X$. Then the map $f:X\times X^*\to F$ ($F=\mathbb{R}$ or $\mathbb{C}$), defined by $f(x,y^*)=y^*(Tx)$ immediately gives us a bilinear form on $X\times X^*$.
Verification: $f(ax_1+bx_2,y^*)=y^*(T(ax_1+bx_2))=y^*(Tax_1)+y^*(Tbx_2)=ay^*(Tx_1)+by^*(Tx_2)=af(x_1,y^*)+bf(x_2,y^*).$
$f(x, ay_1^*+by_2^*)=(ay_1^*+by_2^*)(Tx)=ay_1^*(Tx)+by_2^*(Tx)=af(x,y_1^*)+bf(x,y_2^*).$
But whenever $X$ is Hilbert, we have the following
$f(x,y^*)=y^*(Tx)$. By Riesz Representation theorem, we have $y^*(\cdot)=\langle \cdot, y \rangle$ for some $y\in X$. Thus, we have $$f(x,y^*)=\langle Tx,y \rangle.$$ Thus, here $f$ becomes conjugate linear in the second component.
The Hilbert space case the particular case of the Banach space case. However, in case of Hilbert space the "bilinear" form becomes conjugate linear.
Why there is such an incompatibility? Am I missing anything?
Thanks in advance.
The Riesz Representation theorem says that the function $$\phi : X \to X^*; \quad x\ \mapsto \langle \cdot,x \rangle$$ is a bijective conjugate-linear map. Thus, for $y^* \in X^*$ we have $y^* = \phi(\phi^{-1}(y^*)) = \langle \cdot, \phi^{-1}(y^*) \rangle$ and $$(\forall x \in X) \quad f(x,y^*) = y^*(Tx) = \langle Tx, \phi^{-1}(y^*) \rangle.$$ Since $\phi^{-1}$ and $\langle Tx,\cdot \rangle$ are both conjugate-linear, $\langle Tx,\cdot \rangle \circ \phi^{-1} = f(x,\cdot)$ is linear, and there is no contradiction.