Showing that $(f\circ g)^*=g^*\circ f^*$

Question

I need to show the following property of adjoints: $(f\circ g)^*=g^*\circ f^*$ using the fact that $\langle v,f^*(w)\rangle =\langle f(v),w\rangle$.

I haven't been able to get very far but what I have so far is:

$$\langle v,(f\circ g)^*(w)\rangle=\langle (f\circ g)(v),w\rangle$$

I know I need to eventually get to $\langle v,(g^*\circ f^*)(w)\rangle$ but I don't know how to deal with the composition

Answer 1

One has: $\langle v,g^*(f^*(w))\rangle=\langle g(v),f^*(w)\rangle=\langle f(g(v)),w\rangle$. Using uniqueness of the adjoint: $$(f\circ g)^*=g^*\circ f^*.$$