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Let $(H,(\cdot,\cdot))$ be a pre-Hilbert space on $\mathbb{K}$. A Bilinear form $B:H\times H\rightarrow\mathbb{K}$ is:

  • bounded if there is a $C>0$ so that, $$|B(x,y)|\le C||x||||y|| \forall x,y\in H$$

  • continuous, if for $x_n \rightarrow x$ and $y_n\rightarrow y$ the following is true: $$B(x_n,y_n)\rightarrow B(x,y)$$ for $n\rightarrow \infty$

i) A Bilinear form is bounded if and only of it is continuous

ii) Is $A:H\rightarrow H$ linear then B(x,y):=(Ax,y) is a Bilinear form B on H. B is continuous if and only if A is continuous. In this case $||A||=\sup_{x,y\neq 0}\frac{|B(x,y)|}{||x||||y||}$

I already proofed i) and why $B(x,y):=(Ax,y)$ is a Bilinear form. But I don't know how to show the rest if ii). The statement seems obvious but I don't know how to calculate the norm of A. The only theorem I know, which helps to calculate the norm, defines the norm as: $||A||:=\sup_{||x||\le1}||Tx||_Y<\infty$. Can someone help me?

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Hints:

  1. $|B(x,y)| \leq \|Ax\| \|y\| \leq \|A\| \|x\| \|y\|$.
  2. $B(x,Ax) = (Ax,Ax) = \|Ax\|^2$.
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  • $\begingroup$ Ok I got the direction: B is continuous so A is continuous. Now I try to show that $(Ax,y)\le C||x||||y||$ so that B is bounded$\leftrightarrow$continuous. Could you help me there? $\endgroup$ – Tobi92sr Nov 30 '17 at 22:59

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