jl
发帖数: 398 | 1 向量 Y = X + E
在 L_2 空间, X 和 E 正交;
|Y|^2 = |X|^2 + |E|^2 ;
有没有哪个定理说:
在L_p, p>2, 如果 |X| >0
|Y|^p > |E|^p
多谢!
|
x******a
发帖数: 6336 | 2 what is 正交 in L_p?
【在 jl 的大作中提到】
: 向量 Y = X + E
: 在 L_2 空间, X 和 E 正交;
: |Y|^2 = |X|^2 + |E|^2 ;
: 有没有哪个定理说:
: 在L_p, p>2, 如果 |X| >0
: |Y|^p > |E|^p
: 多谢!
:
|
jl
发帖数: 398 | 3 The inner product is zero in L2 space.
【在 x******a 的大作中提到】
: what is 正交 in L_p?
|
l********e
发帖数: 3632 | 4 你没有想过？
Y=1
X=-1
E=2
你说的不是显然不成立吗？
【在 jl 的大作中提到】
: 向量 Y = X + E
: 在 L_2 空间, X 和 E 正交;
: |Y|^2 = |X|^2 + |E|^2 ;
: 有没有哪个定理说:
: 在L_p, p>2, 如果 |X| >0
: |Y|^p > |E|^p
: 多谢!
:
|
jl
发帖数: 398 | 5 X, E are linear dependent in your case.
【在 l********e 的大作中提到】
: 你没有想过？
: Y=1
: X=-1
: E=2
: 你说的不是显然不成立吗？
|
l********e
发帖数: 3632 | 6 you didn't get the point and you didn't specify you condition on X and E in
your question.
if you still insist, simply let E=2+i and X=-1-i
【在 jl 的大作中提到】
: X, E are linear dependent in your case.
|
n***p
发帖数: 7668 | 7 The answer is no. Here is an counterexample.
Consider the 2D Euclidean space and let
X =(2,1) and E=(-a, 2a). Then X is perpendicular to E.
Then Y=(2-a, 1+2a). For p=4,
|Y|_4^4 = (2-a)^4 + (1+2a)^4
= 17 - 24a + C_1 a^2 +C_2 a^3 +C_3 a^4
<17 = |X|_4^4
if we choose a>0 to be small enough.
The same idea can be applied to construct counterexamples
for other spaces like L_p function spaces for p=4. I choose
p=4 because I can easily calculate the norms explicitly.
For other values of p, we can use taylor expansions.
【在 jl 的大作中提到】
: 向量 Y = X + E
: 在 L_2 空间, X 和 E 正交;
: |Y|^2 = |X|^2 + |E|^2 ;
: 有没有哪个定理说:
: 在L_p, p>2, 如果 |X| >0
: |Y|^p > |E|^p
: 多谢!
:
|
n***p
发帖数: 7668 | 8 I don't get the point either and appreciate it if you may
clarify further. To me I don't see in what definition
E=2+i and X=-1-i can be orthogonal.
in
【在 l********e 的大作中提到】
: you didn't get the point and you didn't specify you condition on X and E in
: your question.
: if you still insist, simply let E=2+i and X=-1-i
|
l********e
发帖数: 3632 | 9 The point is there is no inner which is compatible with the norm for the L^p
(p>2) space. They are only Banach spaces, so it does not make any sense to
talk about orthogonality in such a space.
So basically there is no restriction on X and E in the decomposition in OP's
question.
【在 n***p 的大作中提到】
: I don't get the point either and appreciate it if you may
: clarify further. To me I don't see in what definition
: E=2+i and X=-1-i can be orthogonal.
:
: in
|
n***p
发帖数: 7668 | 10 Since L^p functions are naturally in L^2, it is OK to talk about
orthogonality of L^p functions in L^2. When talking about functions
in L^p\cap L^2, there are naturally two norms, one from L^p, and
another from L^2. It is OK to discuss the relation between these
two norms.
^p
to
's
【在 l********e 的大作中提到】
: The point is there is no inner which is compatible with the norm for the L^p
: (p>2) space. They are only Banach spaces, so it does not make any sense to
: talk about orthogonality in such a space.
: So basically there is no restriction on X and E in the decomposition in OP's
: question.
|
l********e
发帖数: 3632 | 11 no, the claim "L^p functions are naturally in L^2" depends on the measure of
the whole space to be finite, which is apparently not assumed by OP.
【在 n***p 的大作中提到】
: Since L^p functions are naturally in L^2, it is OK to talk about
: orthogonality of L^p functions in L^2. When talking about functions
: in L^p\cap L^2, there are naturally two norms, one from L^p, and
: another from L^2. It is OK to discuss the relation between these
: two norms.
:
: ^p
: to
: 's
|
n***p
发帖数: 7668 | 12 Come on. Apparently OP implicitly made such assumptions.
of
【在 l********e 的大作中提到】
: no, the claim "L^p functions are naturally in L^2" depends on the measure of
: the whole space to be finite, which is apparently not assumed by OP.
|
l********e
发帖数: 3632 | 13 哪里显然？这里毫无上下文你也能分析出这个条件，算你牛。
我都怀疑LZ是指不知道这个条件。
【在 n***p 的大作中提到】
: Come on. Apparently OP implicitly made such assumptions.
:
: of
|
