3
$\begingroup$

I'm looking for some simple proof of the following consequence of the "strengthened" Cauchy-Schwarz inequality:

Let $\mathcal{H}$ be a real Hilbert space such that $\mathcal{H}=\mathcal{V}\oplus\mathcal{W}$ with an inner product $(\cdot,\cdot)$ and the induced norm $\|\cdot\|$. Let $\alpha\in[0,1)$ be the smallest constant such that $$\tag{1} (v,w)\leq\alpha\|v\|\|w\|\quad\forall v\in\mathcal{V},\;w\in\mathcal{W}. $$ Then for any $v\in\mathcal{V}$, $$\tag{2} \|v\|\leq\frac{1}{\sqrt{1-\alpha^2}}\inf_{w\in\mathcal{W}}\|v+w\|. $$

Obviously, (2) is equivalent to $$\tag{3} (1-\alpha^2)(v,v)\leq (v+w,v+w) \quad\Leftrightarrow\quad 0\leq\alpha^2(v,v)+2(v,w)+(w,w) $$ for any $v\in\mathcal{V}$ and $w\in\mathcal{W}$.

I'm however not sure how to verify that (3) holds using (1).

Thanks in advance.

$\endgroup$
3
$\begingroup$

$$\alpha^2(v,v) + 2(v,w) + (w,w) \geqslant \alpha^2(v,v) - 2\alpha\lVert v\rVert\,\lVert w\rVert + (w,w) = \left(\alpha \lVert v\rVert - \lVert w\rVert\right)^2 \geqslant 0,$$

since by $(1)$, we have $(v,w) \geqslant - \alpha\lVert v\rVert\,\lVert w\rVert$.

$\endgroup$
  • 1
    $\begingroup$ So there is the missing step. Thank you. $\endgroup$ – Algebraic Pavel May 14 '14 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.