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!
Best Answer
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 $\|F_{y}\|\leq\|y\|$.