Inner product space and dual space

I noticed there were many questions on this. If anyone could link me to a post where my question is answered, I'd greatly appreciate. Otherwise, any assistance here would also be greatly appreciate it.

An inner product space is a vector space $$V$$ over $$\mathbb{R}$$ with a bilinear, symmetric, non-degenerate form $$(\cdot ,\cdot):V \times V \rightarrow \mathbb{R}$$ with associated norm $$||x||=\sqrt{(x,x)}$$. For every $$y\in V$$ define a function $$F_y:V\rightarrow \mathbb{R}$$ by $$F_y(x)=(x,y)$$. Show that $$F_y\in V*$$ and $$||F_y||=||y||$$.

My attempt:

$$F_y$$ will inherit the linearity from $$(\cdot,\cdot)$$. The inner product is continuous, hence $$F_y \in V^*$$.

I don't really know how to show that $$||F_y||=||y||$$. I know that $$F_y$$ is linear, so $$F_y(x)=(x,y)$$ by linearity $$F_y=y$$ and then $$||F_y||=||y||$$ but that does not seem correct at all.

I ask for help especially with showing $$||F_y||=||y||$$. I am lost there.

Thanks!

Take $$x=y/\|y\|$$ then $$\|x\|=1$$ and $$\|F_{y}\|=\sup_{\|x\|=1}|F_{y}(x)|\geq|F_{y}(x)|=(y/\|y\|,y)=\|y\|^{2}/\|y\|=\|y\|$$.
On the other hand, Cauchy-Schwarz inequality gives that $$|F_{y}(x)|=|(x,y)|\leq\|x\|\|y\|$$ and hence $$\|F_{y}\|\leq\|y\|$$.
• why are you taking the sup? is it so that you have that $\geq$ inequality guaranteed? and when you apply Cauchy-Schwarz, why can you can from $|(x,y)|\leq ||x|| ||y||$ to $||F_y||\leq ||y||$. Thanks for the answer! – Schach21 Dec 20 '19 at 0:29