The correspondence $T:H\rightarrow H\mbox{*}$ by $y\mapsto f_y$, and $y\in H$ and $T$ is a conjugate-linear isometry. We call $H\mbox{*}$ the dual space. We know that something is conjugate-linear if $$T(\lambda u+\mu v)=\bar{\lambda} Tu+\bar{\mu} Tv$$ for all $u\in X,v\in Y$ and $\lambda,\mu\in\mathbb{F}$. And isometry being $||Tx||=||x||$
We want to show that $$\langle Tx,Ty\rangle_1=\langle y,x\rangle$$ defines an inner product on $H\mbox{*}$ that induces a norm on $H\mbox{*}$.
I know that I am supposed to show that $$||x||=0\iff x=0$$ $$\text{Triangle Inequality}$$ $$||\alpha x||=|\alpha|||x||$$ But I am struggling with the exact steps to do this.
I will post what I have in a little if you would be more open to checking what I have.
This is what I was thinking but for some reason I just felt like it wasn't quite right.
[(a)][$||x||=0 \iff x=0$]: \begin{align*} ||Tx||&=0\\ ||Tx||^2&=\\ \langle Tx,Tx\rangle_1&=\\ \langle x,x\rangle&=0\iff x=0\\ \end{align*} [(b)][$||\alpha x||=|\alpha|||x||$]: \begin{align*} ||T\alpha x||&=||\bar{\alpha}Tx||\\ &=||\bar{\alpha} x||\\ &=|\alpha|||x||\\ &=|\alpha|||Tx|| \end{align*} [(c)][Triangle inequality]: \begin{align*} ||Tx+Ty||&=||x+y||\\ &\leq||x||+||y||\\ &=||Tx||+||Ty|| \end{align*}
As already said you haven't proved that $\langle\cdot,\cdot\rangle_1$ defines an inner product. I will give you some hints: $$\langle f,g\rangle_1 := \langle T^{-1}g,T^{-1}f\rangle$$ for every $f,g \in H^*$. Prove that: $$\langle f,g\rangle_1 = \overline{\langle g,f\rangle}_1$$ $$\langle \mu f + \lambda h,g\rangle_1 = \mu \langle f,g\rangle_1 + \lambda \langle h,g\rangle_1$$ $$\langle f,f\rangle_1 \geq 0$$ $$\langle f,f\rangle_1 = 0 \; \text{iff} \; f=0$$ all follows from the properties of $\langle \cdot,\cdot\rangle$