Show that in an inner product space

• April 1st 2011, 11:46 AM
raed
Show that in an inner product space
Dear Colleagues,

Could you please help me in solving the following problem:
Show that in an inner product space, $x\bot y$ if and only if $||x+\alpha y||\geq ||x||$ for all scalars $\alpha$.

Remark: I have already proved that $x\bot y$ implies that $||x+\alpha y||\geq ||x||$ for all scalars $\alpha$, it remains the converse.

Regards,

Raed.
• April 1st 2011, 12:22 PM
girdav
Take the squares and expand the inner product.
• April 1st 2011, 12:51 PM
raed
Thank you very much for your reply, I have already done that and the result is
$2Re(\alpha {\bar} \langle x,y \rangle) + |\alpha|^{2}\langle y,y \rangle \geq 0$, where $Re$ denotes the real part of a complex number.

But what then?
• April 1st 2011, 12:53 PM
girdav
You can divide by $|\alpha|^2$.