Bilinear Form in Definition of Adjoint Operator

I'm working through some material in Analysis Now by Pedersen, and I'm a little confused about how the adjoint to an operator is defined. Pederson defines it by:

If $X$ and $Y$ are normed spaces and $T\in \mathcal{B}(X,Y)$, then we define $T^*:Y^*\rightarrow X^*$ by the equation $\langle x,T^*\phi\rangle = \langle Tx,\phi\rangle$, where $x\in X$ and $\phi\in Y^*.$

I'm a little confused as to what exactly our bilinear form is here. Is the adjoint defined with respect to a particular one? Pederson above uses one where $\langle x,\phi\rangle = \phi (x)$, but I've looked around at a couple other sources and the definition generally looks like the above one without introducing a specified bilinear form at all. And in the special case we're in a Hilbert space I assume we're just considering the inner product?

• It is the standard duality on a normed space and its dual. It is pretty ... standard. – user251257 Feb 24 '16 at 1:54
• @user251257 I assumed there was something standard about it, but I don't know what the standard is. Could you expand your answer? – Dizzy Feb 24 '16 at 2:00

There is nothing special about the bilinear notation. You may as well just write $$y^*\left(Tx\right) = (T^*y^*)(x)$$ in place of $$\langle Tx,y^* \rangle = \langle x,T^*y^*\rangle.$$ But then you lose insight on the symmetry between Banach spaces and their dual spaces without the bilinear notation, especially the fact that $x$ acts like a continuous linear functional on $X^*$.
Suppose $H$ is a Hilbert space, and $T\in \mathscr{B}(H)$, then the adjoint of $T$ is defined by $$(Tx,y) = (x,T^*y).$$ It is slightly different from the Banach adjoint of $T$, because we defined the Hilbert adjoint on $H$ instead of $H^*$.