# Linear Functional

• Apr 22nd 2010, 12:09 PM
ejgmath
Linear Functional
I need to show that a linear functional $\displaystyle \alpha_v$ on a Hilbert space $\displaystyle H$ defined by $\displaystyle \alpha_v(w)=<w,v>$ has operator norm $\displaystyle \|\alpha_v\|_{op}=\|v\|$.

I have already shown by the Cauchy Schwarz inequality that $\displaystyle \|\alpha_v\|_{op}\leq\|v\|$.

I think now I need to show that $\displaystyle \|\alpha_v\|_{op}\geq\|v\|$.

Any help would be great, Thanks
• Apr 22nd 2010, 12:27 PM
Focus
Quote:

Originally Posted by ejgmath
I need to show that a linear functional $\displaystyle \alpha_v$ on a Hilbert space $\displaystyle H$ defined by $\displaystyle \alpha_v(w)=<w,v>$ has operator norm $\displaystyle \|\alpha_v\|_{op}=\|v\|$.

I have already shown by the Cauchy Schwarz inequality that $\displaystyle \|\alpha_v\|_{op}\leq\|v\|$.

I think now I need to show that $\displaystyle \|\alpha_v\|_{op}\geq\|v\|$.

Any help would be great, Thanks

Tip, plug in $\displaystyle v/||v||$.
• Apr 22nd 2010, 12:41 PM
ejgmath
Into where, can you elabourate?
• Apr 22nd 2010, 01:47 PM
Focus
Quote:

Originally Posted by ejgmath
Into where, can you elabourate?

Into $\displaystyle \alpha_v$. What you want is $\displaystyle \alpha_v(x)=||v||$ for some x with $\displaystyle ||x|| \leq 1$, which precisely shows the inequality you are looking for (as $\displaystyle ||\alpha_v||:=\sup \{|\alpha_v(x)|: ||x|| \leq 1 \}$ ).