Linear Functional

**ejgmath** · April 22nd 2010, 12:09 PM

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

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

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

Any help would be great, Thanks

**Focus** · April 22nd 2010, 12:27 PM

Originally Posted by ejgmath

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

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

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

Any help would be great, Thanks

Tip, plug in $v/||v||$ .

**ejgmath** · April 22nd 2010, 12:41 PM

Into where, can you elabourate?

**Focus** · April 22nd 2010, 01:47 PM

Originally Posted by ejgmath

Into where, can you elabourate?

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

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