Let $E$ and $F$ finite vector space. I know that if $u\in \mathcal L(E,F)$ we define the adjoint $u^*$ of $u$ as $u^*(f)=f(u)$, i.e. $u^*\in \mathcal L(F^*,E^*)$. It's written in my book that $$\left<u(x),f\right>_{F,F^*}=\left<x,u^*(f)\right>_{E,E^*},$$ for all $f\in F^*$ and all $x\in E$.
I would like to make a similarity with the construction of the adjoint in a inner space. I recall that if $(E,\left<\cdot ,\cdot \right>)$ is a inner space of finite dimension, and $T\in \mathcal L(E)$, then for $v\in E$ fixed, $$\varphi (u)=\left<T(u),v\right>,$$ is a linear form, and thus, there is a unique $v(T)$ s.t. $$\left< T(u),v\right>=\varphi(u)=\left<u,v(T)\right> .$$ We denote $v(T):= T^*v$. The application $T^*$ is well defined since for all $v\in V$, there is a unique $w\in V$ s.t. $w=T^*v$.
Now, I'm trying to reproduce this construction.
First try If I define $$\varphi(x):=\left<u(x),f\right>_{F,F^*}$$ then $\varphi\in E^*$. I don't know theorem in this general case that tels me that there is a unique $u^*(f)\in E^*$ s.t. $\varphi(x)=\left<x,u^*(f)\right>$.
Second try So I tried to use that bijection $$\Psi :F\to F^{**}$$ defined as $$\Psi(x)(f)=\left<f,x\right>:= f(x),\quad f\in F^*.$$
Unfortunately $x\longmapsto \left<u(x),f\right>$ is an element of $E^*$ and not of $E^{**}$, so I can't conclude.
So is there a similarity between the definition of the adjoint of $u$ when we are in inner space and when we are in more general spaces ?