Mixing
Show $langle x-Px,y-Pxrangle leq 0$ for $Px$ the projection of $x$ onto $C$ convex, and any $y in C$.
$begingroup$
So far I did the following:
Write $x=w+Px$ for $w$ st. $Pw=0$. Then $langle x-Px,y-Pxrangle=langle w,y-Px rangle$. From here I can intuitively see that is non-positive since 'the arrow' from $Px$ to $y$ is pointed inside $C$ and $w$ is at a right angle to the 'surface' of $C$ at $Px$ but obviously this is not a proof.
Can anybody give me a hint?
trigonometry convex-analysis inner-product-space
asked Jan 22 at 13:07
DanDan
958
$endgroup$
add a comment |
$begingroup$
So far I did the following:
Write $x=w+Px$ for $w$ st. $Pw=0$. Then $langle x-Px,y-Pxrangle=langle w,y-Px rangle$. From here I can intuitively see that is non-positive since 'the arrow' from $Px$ to $y$ is pointed inside $C$ and $w$ is at a right angle to the 'surface' of $C$ at $Px$ but obviously this is not a proof.
Can anybody give me a hint?
trigonometry convex-analysis inner-product-space
asked Jan 22 at 13:07
DanDan
958
$endgroup$
1
$begingroup$
Write that $t in [0,1] longmapsto |x-(1-t)Px-ty|^2$ is minimal at $t=0$.
$endgroup$
– Mindlack
Jan 22 at 13:10
$begingroup$
Great tip @Mindlack, $||x-(1-t)Px-ty||^2=c-2t langle x-Px,y-Px rangle$ with $c geq 0$ and this function is increasing in $t$ implying the result.
$endgroup$
– Dan
Jan 22 at 15:36
add a comment |
$begingroup$
So far I did the following:
Write $x=w+Px$ for $w$ st. $Pw=0$. Then $langle x-Px,y-Pxrangle=langle w,y-Px rangle$. From here I can intuitively see that is non-positive since 'the arrow' from $Px$ to $y$ is pointed inside $C$ and $w$ is at a right angle to the 'surface' of $C$ at $Px$ but obviously this is not a proof.
Can anybody give me a hint?
trigonometry convex-analysis inner-product-space
asked Jan 22 at 13:07
DanDan
958
$endgroup$
So far I did the following:
Write $x=w+Px$ for $w$ st. $Pw=0$. Then $langle x-Px,y-Pxrangle=langle w,y-Px rangle$. From here I can intuitively see that is non-positive since 'the arrow' from $Px$ to $y$ is pointed inside $C$ and $w$ is at a right angle to the 'surface' of $C$ at $Px$ but obviously this is not a proof.
Can anybody give me a hint?
trigonometry convex-analysis inner-product-space
trigonometry convex-analysis inner-product-space
asked Jan 22 at 13:07
DanDan
958
asked Jan 22 at 13:07
DanDan
958
asked Jan 22 at 13:07
DanDan
958
asked Jan 22 at 13:07
DanDan
958
asked Jan 22 at 13:07
DanDan
958
958
1
$begingroup$
Write that $t in [0,1] longmapsto |x-(1-t)Px-ty|^2$ is minimal at $t=0$.
$endgroup$
– Mindlack
Jan 22 at 13:10
$begingroup$
Great tip @Mindlack, $||x-(1-t)Px-ty||^2=c-2t langle x-Px,y-Px rangle$ with $c geq 0$ and this function is increasing in $t$ implying the result.
$endgroup$
– Dan
Jan 22 at 15:36
add a comment |
1
$begingroup$
Write that $t in [0,1] longmapsto |x-(1-t)Px-ty|^2$ is minimal at $t=0$.
$endgroup$
– Mindlack
Jan 22 at 13:10
$begingroup$
Great tip @Mindlack, $||x-(1-t)Px-ty||^2=c-2t langle x-Px,y-Px rangle$ with $c geq 0$ and this function is increasing in $t$ implying the result.
$endgroup$
– Dan
Jan 22 at 15:36
1
1
$begingroup$
Write that $t in [0,1] longmapsto |x-(1-t)Px-ty|^2$ is minimal at $t=0$.
$endgroup$
– Mindlack
Jan 22 at 13:10
$begingroup$
Write that $t in [0,1] longmapsto |x-(1-t)Px-ty|^2$ is minimal at $t=0$.
$endgroup$
– Mindlack
Jan 22 at 13:10
$begingroup$
Great tip @Mindlack, $||x-(1-t)Px-ty||^2=c-2t langle x-Px,y-Px rangle$ with $c geq 0$ and this function is increasing in $t$ implying the result.
$endgroup$
– Dan
Jan 22 at 15:36
$begingroup$
Great tip @Mindlack, $||x-(1-t)Px-ty||^2=c-2t langle x-Px,y-Px rangle$ with $c geq 0$ and this function is increasing in $t$ implying the result.
$endgroup$
– Dan
Jan 22 at 15:36
add a comment |
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1
$begingroup$
Write that $t in [0,1] longmapsto |x-(1-t)Px-ty|^2$ is minimal at $t=0$.
$endgroup$
– Mindlack
Jan 22 at 13:10
$begingroup$
Great tip @Mindlack, $||x-(1-t)Px-ty||^2=c-2t langle x-Px,y-Px rangle$ with $c geq 0$ and this function is increasing in $t$ implying the result.
$endgroup$
– Dan
Jan 22 at 15:36