This is the following definition we are using for Adjoing Operator on normed vectorial spaces:
Let $A \in \mathcal{L}(E,F)$, the adjoint of $A$ is the only bounded operator $A^*:F^* \rightarrow E^*$ s.t,
\begin{equation}g(Ax) = (A^*g)(x) \ \ \ \ \ \ \ \forall x\in E, \ \forall g\in F^*.\end{equation}
And with this definition we wouldn't need any inner product, but when trying to find adjoint of operators I'm finding trouble, both with finding its adjoint and to understand this equivalence with spaces with inner product.
Example:
1) $A:L^2([0,1])\rightarrow L^2([0,1])$, $A_x(t) = \int_{0}^{t} x(s) ds$
Using inner product we have
\begin{equation} <g,Ax> = \int_0^1 g(t)\int_0^t x(s) ds \ dt = \int_0^1 x(s) \int_s^1 g(t) dt \ ds = <A^*g,x> \end{equation} giving us, $A^*g (s) = \int_s^1 g$, but i can't compute this operator without using this inner product and going via the definition i have (as composition), and I'm having trouble understanding how they would be equivalent, but in this case it's okay as I have an inner product to rely on (even the exercise should be solved without it as far as i know).
2)$A: l^1(\mathbb{N}) \rightarrow l^1(\mathbb{N})$, $A_x =(x_1, ... ,x_n, 0,0,0,...)$ with fixed n,
in this case if i were to use the $l^2$ inner product i would have \begin{equation} <g,Ax> = \sum y_i x_i = <A^*g,x> \end{equation} and then $A$ would be self adjoint, but here i can't even tell if this is correct as I'm "borrowing" the inner product.
Any help to clarify these ideas would be apreciated!