Mixing
Proving inequality involving scalar product on $mathbb{R}^n$
$begingroup$
Suppose $n,mgeq 1$, $Asubset mathbb{R}^m$ is compact and $f:mathbb{R}^ntimes Ato mathbb{R}^n$ is a continuous function satisfying $$leftlVert f(x_1,a)-f(x_2,a)rightrVert leq LleftlVert x_1-x_2rightrVert qquad (L>0)$$ for all $x_1,x_2 in mathbb{R}^n$ and $ain A$.
I have to prove that $$f(y, a)cdot yleq K(1+{leftlVert yrightrVert}^2)qquad [1]$$ for all $yin mathbb{R}^n$ and $ain A$ where $$K=L+sup_{ain A}{leftlVert f(0,a)rightrVert}$$ and $cdot$ is the scalar product on $mathbb{R}^n$.
For all $a in A$ and $y in mathbb{R}^n$ we have, by the Cauchy-Schwarz inequality and the Lipschitz condition on $f$, $$[f(y,a)-f(0,a)]cdot yleq vert[f(y,a)-f(0,a)]cdot y|leq leftlVert f(y,a)-f(0,a)rightrVertleftlVert yrightrVertleq L{leftlVert yrightrVert}^2,$$ and therefore $$f(y,a)cdot yleq f(0,a)cdot y +L{leftlVert yrightrVert}^2leqleftlVert f(0,a)rightrVertleftlVert yrightrVert+L{leftlVert yrightrVert}^2 \ leq sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2. $$ Now the author concludes saying that this implies $[1]$, but I really don't understand why.
Any hint?
Thanks a lot in advance
normed-spaces inner-product-space supremum-and-infimum
edited Jan 27 at 22:17
max_zorn
3,43061429
asked Jan 27 at 21:35
eleguitareleguitar
140114
$endgroup$
add a comment |
$begingroup$
Suppose $n,mgeq 1$, $Asubset mathbb{R}^m$ is compact and $f:mathbb{R}^ntimes Ato mathbb{R}^n$ is a continuous function satisfying $$leftlVert f(x_1,a)-f(x_2,a)rightrVert leq LleftlVert x_1-x_2rightrVert qquad (L>0)$$ for all $x_1,x_2 in mathbb{R}^n$ and $ain A$.
I have to prove that $$f(y, a)cdot yleq K(1+{leftlVert yrightrVert}^2)qquad [1]$$ for all $yin mathbb{R}^n$ and $ain A$ where $$K=L+sup_{ain A}{leftlVert f(0,a)rightrVert}$$ and $cdot$ is the scalar product on $mathbb{R}^n$.
For all $a in A$ and $y in mathbb{R}^n$ we have, by the Cauchy-Schwarz inequality and the Lipschitz condition on $f$, $$[f(y,a)-f(0,a)]cdot yleq vert[f(y,a)-f(0,a)]cdot y|leq leftlVert f(y,a)-f(0,a)rightrVertleftlVert yrightrVertleq L{leftlVert yrightrVert}^2,$$ and therefore $$f(y,a)cdot yleq f(0,a)cdot y +L{leftlVert yrightrVert}^2leqleftlVert f(0,a)rightrVertleftlVert yrightrVert+L{leftlVert yrightrVert}^2 \ leq sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2. $$ Now the author concludes saying that this implies $[1]$, but I really don't understand why.
Any hint?
Thanks a lot in advance
normed-spaces inner-product-space supremum-and-infimum
edited Jan 27 at 22:17
max_zorn
3,43061429
asked Jan 27 at 21:35
eleguitareleguitar
140114
$endgroup$
add a comment |
$begingroup$
Suppose $n,mgeq 1$, $Asubset mathbb{R}^m$ is compact and $f:mathbb{R}^ntimes Ato mathbb{R}^n$ is a continuous function satisfying $$leftlVert f(x_1,a)-f(x_2,a)rightrVert leq LleftlVert x_1-x_2rightrVert qquad (L>0)$$ for all $x_1,x_2 in mathbb{R}^n$ and $ain A$.
I have to prove that $$f(y, a)cdot yleq K(1+{leftlVert yrightrVert}^2)qquad [1]$$ for all $yin mathbb{R}^n$ and $ain A$ where $$K=L+sup_{ain A}{leftlVert f(0,a)rightrVert}$$ and $cdot$ is the scalar product on $mathbb{R}^n$.
For all $a in A$ and $y in mathbb{R}^n$ we have, by the Cauchy-Schwarz inequality and the Lipschitz condition on $f$, $$[f(y,a)-f(0,a)]cdot yleq vert[f(y,a)-f(0,a)]cdot y|leq leftlVert f(y,a)-f(0,a)rightrVertleftlVert yrightrVertleq L{leftlVert yrightrVert}^2,$$ and therefore $$f(y,a)cdot yleq f(0,a)cdot y +L{leftlVert yrightrVert}^2leqleftlVert f(0,a)rightrVertleftlVert yrightrVert+L{leftlVert yrightrVert}^2 \ leq sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2. $$ Now the author concludes saying that this implies $[1]$, but I really don't understand why.
Any hint?
Thanks a lot in advance
normed-spaces inner-product-space supremum-and-infimum
edited Jan 27 at 22:17
max_zorn
3,43061429
asked Jan 27 at 21:35
eleguitareleguitar
140114
$endgroup$
Suppose $n,mgeq 1$, $Asubset mathbb{R}^m$ is compact and $f:mathbb{R}^ntimes Ato mathbb{R}^n$ is a continuous function satisfying $$leftlVert f(x_1,a)-f(x_2,a)rightrVert leq LleftlVert x_1-x_2rightrVert qquad (L>0)$$ for all $x_1,x_2 in mathbb{R}^n$ and $ain A$.
I have to prove that $$f(y, a)cdot yleq K(1+{leftlVert yrightrVert}^2)qquad [1]$$ for all $yin mathbb{R}^n$ and $ain A$ where $$K=L+sup_{ain A}{leftlVert f(0,a)rightrVert}$$ and $cdot$ is the scalar product on $mathbb{R}^n$.
For all $a in A$ and $y in mathbb{R}^n$ we have, by the Cauchy-Schwarz inequality and the Lipschitz condition on $f$, $$[f(y,a)-f(0,a)]cdot yleq vert[f(y,a)-f(0,a)]cdot y|leq leftlVert f(y,a)-f(0,a)rightrVertleftlVert yrightrVertleq L{leftlVert yrightrVert}^2,$$ and therefore $$f(y,a)cdot yleq f(0,a)cdot y +L{leftlVert yrightrVert}^2leqleftlVert f(0,a)rightrVertleftlVert yrightrVert+L{leftlVert yrightrVert}^2 \ leq sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2. $$ Now the author concludes saying that this implies $[1]$, but I really don't understand why.
Any hint?
Thanks a lot in advance
normed-spaces inner-product-space supremum-and-infimum
normed-spaces inner-product-space supremum-and-infimum
edited Jan 27 at 22:17
max_zorn
3,43061429
asked Jan 27 at 21:35
eleguitareleguitar
140114
edited Jan 27 at 22:17
max_zorn
3,43061429
asked Jan 27 at 21:35
eleguitareleguitar
140114
edited Jan 27 at 22:17
max_zorn
3,43061429
edited Jan 27 at 22:17
max_zorn
3,43061429
edited Jan 27 at 22:17
max_zorn
3,43061429
3,43061429
asked Jan 27 at 21:35
eleguitareleguitar
140114
asked Jan 27 at 21:35
eleguitareleguitar
140114
asked Jan 27 at 21:35
eleguitareleguitar
140114
140114
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
When $leftlVert yrightrVertge 1$, we find that
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K|y|^2le K(|y|^2+1).
$$ When $leftlVert yrightrVertle 1$, we can observe
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le Kle K(|y|^2+1).
$$ Thus we have
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K(|y|^2+1).
$$ as desired.
answered Jan 27 at 21:47
SongSong
18.5k21651
$endgroup$
add a comment |
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1 Answer
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$begingroup$
When $leftlVert yrightrVertge 1$, we find that
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K|y|^2le K(|y|^2+1).
$$ When $leftlVert yrightrVertle 1$, we can observe
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le Kle K(|y|^2+1).
$$ Thus we have
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K(|y|^2+1).
$$ as desired.
answered Jan 27 at 21:47
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
When $leftlVert yrightrVertge 1$, we find that
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K|y|^2le K(|y|^2+1).
$$ When $leftlVert yrightrVertle 1$, we can observe
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le Kle K(|y|^2+1).
$$ Thus we have
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K(|y|^2+1).
$$ as desired.
answered Jan 27 at 21:47
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
When $leftlVert yrightrVertge 1$, we find that
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K|y|^2le K(|y|^2+1).
$$ When $leftlVert yrightrVertle 1$, we can observe
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le Kle K(|y|^2+1).
$$ Thus we have
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K(|y|^2+1).
$$ as desired.
answered Jan 27 at 21:47
SongSong
18.5k21651
$endgroup$
When $leftlVert yrightrVertge 1$, we find that
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K|y|^2le K(|y|^2+1).
$$ When $leftlVert yrightrVertle 1$, we can observe
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le Kle K(|y|^2+1).
$$ Thus we have
$$
sup_{a in A}leftlVert f(0,a)rightrVert leftlVert yrightrVert +L{leftlVert yrightrVert}^2le K(|y|^2+1).
$$ as desired.
answered Jan 27 at 21:47
SongSong
18.5k21651
edited Jan 27 at 21:55
answered Jan 27 at 21:47
SongSong
18.5k21651
answered Jan 27 at 21:47
SongSong
18.5k21651
answered Jan 27 at 21:47
SongSong
18.5k21651
18.5k21651
add a comment |
add a comment |
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