Prove norm on any Hilbert space is strictly convex The 2019 Stack Overflow Developer Survey Results Are InHow can I prove that every inner product space is strictly convex normed space?Uniformly convex implies strictly convexProve vectorspace of bounded functions with supremum-norm is complete and no hilbert spaceComposition of projections has a fixed point in a Hilbert spaceStrictly convex setIs Jensen's inequality an iff condition on convex functions?Stronger Condition for Strict Convexity?Is $C([0,1])$ strictly convex?Strictly Monotonic, Continuous Sub-linear functions with F(1)=1; Any real difference between these and linear functions?Prove that a set $ f:lVert frVert_inftyleq 1 $ is not strictly convexCan we define a norm such that the space of all infinite sequences is a Hilbert space?Do $|f'|$ and $f$ have the same minimisers for strictly convex functions?
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Prove norm on any Hilbert space is strictly convex
The 2019 Stack Overflow Developer Survey Results Are InHow can I prove that every inner product space is strictly convex normed space?Uniformly convex implies strictly convexProve vectorspace of bounded functions with supremum-norm is complete and no hilbert spaceComposition of projections has a fixed point in a Hilbert spaceStrictly convex setIs Jensen's inequality an iff condition on convex functions?Stronger Condition for Strict Convexity?Is $C([0,1])$ strictly convex?Strictly Monotonic, Continuous Sub-linear functions with F(1)=1; Any real difference between these and linear functions?Prove that a set $ f:lVert frVert_inftyleq 1 $ is not strictly convexCan we define a norm such that the space of all infinite sequences is a Hilbert space?Do $|f'|$ and $f$ have the same minimisers for strictly convex functions?
$begingroup$
I've seen in lecture notes that norm on any Hilbert space is strictly convex means
"$|x|=|y|=1, quad|x+y|=2 Rightarrow x=y$"
But why this means strict convexity? I thought strict convexity means
$forall x_1 neq x_2 in H, forall t in(0,1) : quad |t x_1+(1-t) x_2|<t |fleft(x_1right)|+(1-t)| fleft(x_2right)|$?
It would be best if you can please give a complete proof of why norm on "any" Hilbert space is strictly convex
convex-analysis hilbert-spaces inner-product-space
asked Apr 8 at 1:28
The RThe R
6710
$endgroup$
add a comment |
$begingroup$
I've seen in lecture notes that norm on any Hilbert space is strictly convex means
"$|x|=|y|=1, quad|x+y|=2 Rightarrow x=y$"
But why this means strict convexity? I thought strict convexity means
$forall x_1 neq x_2 in H, forall t in(0,1) : quad |t x_1+(1-t) x_2|<t |fleft(x_1right)|+(1-t)| fleft(x_2right)|$?
It would be best if you can please give a complete proof of why norm on "any" Hilbert space is strictly convex
convex-analysis hilbert-spaces inner-product-space
asked Apr 8 at 1:28
The RThe R
6710
$endgroup$
$begingroup$
Perhaps it would help you to think of strict convexity in terms of an inequality: if $x neq y$ are two points for which $||x||=1=||y||$, it follows that $$||(x+y)/2||<1$$
$endgroup$
– TM Gallagher
Apr 8 at 1:50
2
$begingroup$
My answer here may help.
$endgroup$
– Theo Bendit
Apr 8 at 2:14
$begingroup$
Possible duplicate: math.stackexchange.com/questions/2133559/…
$endgroup$
– avs
Apr 8 at 5:21
2
$begingroup$
@avs A helpful link, but not really a duplicate. The asker is more confused about how the (seemingly) weaker form of strict convexity implies the stronger form.
$endgroup$
– Theo Bendit
Apr 8 at 10:08
add a comment |
$begingroup$
I've seen in lecture notes that norm on any Hilbert space is strictly convex means
"$|x|=|y|=1, quad|x+y|=2 Rightarrow x=y$"
But why this means strict convexity? I thought strict convexity means
$forall x_1 neq x_2 in H, forall t in(0,1) : quad |t x_1+(1-t) x_2|<t |fleft(x_1right)|+(1-t)| fleft(x_2right)|$?
It would be best if you can please give a complete proof of why norm on "any" Hilbert space is strictly convex
convex-analysis hilbert-spaces inner-product-space
asked Apr 8 at 1:28
The RThe R
6710
$endgroup$
I've seen in lecture notes that norm on any Hilbert space is strictly convex means
"$|x|=|y|=1, quad|x+y|=2 Rightarrow x=y$"
But why this means strict convexity? I thought strict convexity means
$forall x_1 neq x_2 in H, forall t in(0,1) : quad |t x_1+(1-t) x_2|<t |fleft(x_1right)|+(1-t)| fleft(x_2right)|$?
It would be best if you can please give a complete proof of why norm on "any" Hilbert space is strictly convex
convex-analysis hilbert-spaces inner-product-space
convex-analysis hilbert-spaces inner-product-space
asked Apr 8 at 1:28
The RThe R
6710
asked Apr 8 at 1:28
The RThe R
6710
asked Apr 8 at 1:28
The RThe R
6710
asked Apr 8 at 1:28
The RThe R
6710
asked Apr 8 at 1:28
The RThe R
6710
6710
$begingroup$
Perhaps it would help you to think of strict convexity in terms of an inequality: if $x neq y$ are two points for which $||x||=1=||y||$, it follows that $$||(x+y)/2||<1$$
$endgroup$
– TM Gallagher
Apr 8 at 1:50
2
$begingroup$
My answer here may help.
$endgroup$
– Theo Bendit
Apr 8 at 2:14
$begingroup$
Possible duplicate: math.stackexchange.com/questions/2133559/…
$endgroup$
– avs
Apr 8 at 5:21
2
$begingroup$
@avs A helpful link, but not really a duplicate. The asker is more confused about how the (seemingly) weaker form of strict convexity implies the stronger form.
$endgroup$
– Theo Bendit
Apr 8 at 10:08
add a comment |
$begingroup$
Perhaps it would help you to think of strict convexity in terms of an inequality: if $x neq y$ are two points for which $||x||=1=||y||$, it follows that $$||(x+y)/2||<1$$
$endgroup$
– TM Gallagher
Apr 8 at 1:50
2
$begingroup$
My answer here may help.
$endgroup$
– Theo Bendit
Apr 8 at 2:14
$begingroup$
Possible duplicate: math.stackexchange.com/questions/2133559/…
$endgroup$
– avs
Apr 8 at 5:21
2
$begingroup$
@avs A helpful link, but not really a duplicate. The asker is more confused about how the (seemingly) weaker form of strict convexity implies the stronger form.
$endgroup$
– Theo Bendit
Apr 8 at 10:08
$begingroup$
Perhaps it would help you to think of strict convexity in terms of an inequality: if $x neq y$ are two points for which $||x||=1=||y||$, it follows that $$||(x+y)/2||<1$$
$endgroup$
– TM Gallagher
Apr 8 at 1:50
$begingroup$
Perhaps it would help you to think of strict convexity in terms of an inequality: if $x neq y$ are two points for which $||x||=1=||y||$, it follows that $$||(x+y)/2||<1$$
$endgroup$
– TM Gallagher
Apr 8 at 1:50
2
2
$begingroup$
My answer here may help.
$endgroup$
– Theo Bendit
Apr 8 at 2:14
$begingroup$
My answer here may help.
$endgroup$
– Theo Bendit
Apr 8 at 2:14
$begingroup$
Possible duplicate: math.stackexchange.com/questions/2133559/…
$endgroup$
– avs
Apr 8 at 5:21
$begingroup$
Possible duplicate: math.stackexchange.com/questions/2133559/…
$endgroup$
– avs
Apr 8 at 5:21
2
2
$begingroup$
@avs A helpful link, but not really a duplicate. The asker is more confused about how the (seemingly) weaker form of strict convexity implies the stronger form.
$endgroup$
– Theo Bendit
Apr 8 at 10:08
$begingroup$
@avs A helpful link, but not really a duplicate. The asker is more confused about how the (seemingly) weaker form of strict convexity implies the stronger form.
$endgroup$
– Theo Bendit
Apr 8 at 10:08
add a comment |
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$begingroup$
Perhaps it would help you to think of strict convexity in terms of an inequality: if $x neq y$ are two points for which $||x||=1=||y||$, it follows that $$||(x+y)/2||<1$$
$endgroup$
– TM Gallagher
Apr 8 at 1:50
2
$begingroup$
My answer here may help.
$endgroup$
– Theo Bendit
Apr 8 at 2:14
$begingroup$
Possible duplicate: math.stackexchange.com/questions/2133559/…
$endgroup$
– avs
Apr 8 at 5:21
2
$begingroup$
@avs A helpful link, but not really a duplicate. The asker is more confused about how the (seemingly) weaker form of strict convexity implies the stronger form.
$endgroup$
– Theo Bendit
Apr 8 at 10:08