# Linear Functional

• April 22nd 2010, 12:09 PM
ejgmath
Linear Functional
I need to show that a linear functional $\alpha_v$ on a Hilbert space $H$ defined by $\alpha_v(w)=$ 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
• April 22nd 2010, 12:27 PM
Focus
Quote:

Originally Posted by ejgmath
I need to show that a linear functional $\alpha_v$ on a Hilbert space $H$ defined by $\alpha_v(w)=$ 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 $v/||v||$.
• April 22nd 2010, 12:41 PM
ejgmath
Into where, can you elabourate?
• April 22nd 2010, 01:47 PM
Focus
Quote:

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 $||\alpha_v||:=\sup \{|\alpha_v(x)|: ||x|| \leq 1 \}$ ).