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.



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 $\|F_{y}\|\leq\|y\|$.

  • $\begingroup$ 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! $\endgroup$ – Schach21 Dec 20 '19 at 0:29
  • 1
    $\begingroup$ Taking the sup is part of the definition of the operator norm. $\endgroup$ – user284331 Dec 20 '19 at 0:31

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