The norm of a bounded linear functional on a Hilbert space is the norm of the vector?

Question

If $L$ is a bounded linear functional on a Hilbert space $H$, then we know that $$Lx=(x,y),\quad \forall x\in H,$$ for some $y\in H$. Is it true that $\|L\|=\|y\|$?

We have by Cauchy-Schwarz that $$|Lx|=|(x,y)|\leq\|x\|\|y\|,$$ so $\|L\|\leq\|y\|$, but what about the other inequality?

score 3 · Accepted Answer · answered Mar 10 '15 at 23:51

3

Yes. Unless $y=0$, we have $$\|L\|=\sup_{\|x\|=1}|(x,y)|\geq|(y/\|y\|,y)|=\|y\|.$$ If $y=0$, then $\|L\|=0=\|y\|$.

answered Mar 10 '15 at 23:51

Spenser

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The norm of a bounded linear functional on a Hilbert space is the norm of the vector?

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