Mixing
Inequality of norms
$begingroup$
Is it obvious that in $mathbb R^n, forall p,q text{ s.t. } p leq q$
$$ vert vert xvert vert_{p} leq vert vert xvert vert_{q}, text{where }vert vert xvert vert_{q} = (sum_0^n x^q)^{frac 1 q }$$
If yes, how to prove it ? I was thinking about the classical inequalities like Minkowski, Holder or Cauchy Schwartz but none seems to work here.
Moreover, is this inequality still true for any norm ? Like the norm $L_p $ and $L_q$ ? (I mean by this $ (int vert f(x)^q vert dx ) ^{frac 1 q } $ )
In other words, is it true for a space of infinite dimension ?
What about the infinite norm ? is it also always bigger than any other one ? (this one I know is false for the case $mathbb R^n $, but it is true for functions spaces like the space of continuous functions...)
P.S. I'm sorry, I have not yet studied functionnal analysis. I'm asking this for my metric space class where we are proving that some spaces are complete, and I'm wondering when is it possible to use the very usefull inequality of norm I was talking about.
I'm asking this because intuitively, it seems right :
real-analysis norm normed-spaces
asked Jan 23 at 18:37
Marine GalantinMarine Galantin
875319
$endgroup$
add a comment |
$begingroup$
Is it obvious that in $mathbb R^n, forall p,q text{ s.t. } p leq q$
$$ vert vert xvert vert_{p} leq vert vert xvert vert_{q}, text{where }vert vert xvert vert_{q} = (sum_0^n x^q)^{frac 1 q }$$
If yes, how to prove it ? I was thinking about the classical inequalities like Minkowski, Holder or Cauchy Schwartz but none seems to work here.
Moreover, is this inequality still true for any norm ? Like the norm $L_p $ and $L_q$ ? (I mean by this $ (int vert f(x)^q vert dx ) ^{frac 1 q } $ )
In other words, is it true for a space of infinite dimension ?
What about the infinite norm ? is it also always bigger than any other one ? (this one I know is false for the case $mathbb R^n $, but it is true for functions spaces like the space of continuous functions...)
P.S. I'm sorry, I have not yet studied functionnal analysis. I'm asking this for my metric space class where we are proving that some spaces are complete, and I'm wondering when is it possible to use the very usefull inequality of norm I was talking about.
I'm asking this because intuitively, it seems right :
real-analysis norm normed-spaces
asked Jan 23 at 18:37
Marine GalantinMarine Galantin
875319
$endgroup$
add a comment |
$begingroup$
Is it obvious that in $mathbb R^n, forall p,q text{ s.t. } p leq q$
$$ vert vert xvert vert_{p} leq vert vert xvert vert_{q}, text{where }vert vert xvert vert_{q} = (sum_0^n x^q)^{frac 1 q }$$
If yes, how to prove it ? I was thinking about the classical inequalities like Minkowski, Holder or Cauchy Schwartz but none seems to work here.
Moreover, is this inequality still true for any norm ? Like the norm $L_p $ and $L_q$ ? (I mean by this $ (int vert f(x)^q vert dx ) ^{frac 1 q } $ )
In other words, is it true for a space of infinite dimension ?
What about the infinite norm ? is it also always bigger than any other one ? (this one I know is false for the case $mathbb R^n $, but it is true for functions spaces like the space of continuous functions...)
P.S. I'm sorry, I have not yet studied functionnal analysis. I'm asking this for my metric space class where we are proving that some spaces are complete, and I'm wondering when is it possible to use the very usefull inequality of norm I was talking about.
I'm asking this because intuitively, it seems right :
real-analysis norm normed-spaces
asked Jan 23 at 18:37
Marine GalantinMarine Galantin
875319
$endgroup$
Is it obvious that in $mathbb R^n, forall p,q text{ s.t. } p leq q$
$$ vert vert xvert vert_{p} leq vert vert xvert vert_{q}, text{where }vert vert xvert vert_{q} = (sum_0^n x^q)^{frac 1 q }$$
If yes, how to prove it ? I was thinking about the classical inequalities like Minkowski, Holder or Cauchy Schwartz but none seems to work here.
Moreover, is this inequality still true for any norm ? Like the norm $L_p $ and $L_q$ ? (I mean by this $ (int vert f(x)^q vert dx ) ^{frac 1 q } $ )
In other words, is it true for a space of infinite dimension ?
What about the infinite norm ? is it also always bigger than any other one ? (this one I know is false for the case $mathbb R^n $, but it is true for functions spaces like the space of continuous functions...)
P.S. I'm sorry, I have not yet studied functionnal analysis. I'm asking this for my metric space class where we are proving that some spaces are complete, and I'm wondering when is it possible to use the very usefull inequality of norm I was talking about.
I'm asking this because intuitively, it seems right :
real-analysis norm normed-spaces
real-analysis norm normed-spaces
asked Jan 23 at 18:37
Marine GalantinMarine Galantin
875319
asked Jan 23 at 18:37
Marine GalantinMarine Galantin
875319
asked Jan 23 at 18:37
Marine GalantinMarine Galantin
875319
asked Jan 23 at 18:37
Marine GalantinMarine Galantin
875319
asked Jan 23 at 18:37
Marine GalantinMarine Galantin
875319
875319
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Actually, direction should be reversed: it holds that
$$
|x|_pge |x|_q,quad 1le p< qleinfty.tag{*}
$$ What your figure shows us is that
$$
B_psubset B_q,quad p< qtag{**}
$$ where $B_r$ is a unit ball in $r$-norm. It is easy to see that $text{(*)}$ is equivalent to $text{(**)}$.
Now, we prove $text{(*)}$. By the homogeneity of norms, it is sufficient to show that
$$
|x|_qle 1
$$ provided that $|x|_p= 1$. By the assumption, we have
$$
|x_i|^ple |x_1|^p+|x_2|^p+cdots +|x_n|^p=1
$$ for all $ile n$. Hence $|x_i|le 1$ and it follows
$$
|x_i|^qle |x_i|^p,quad i=1,2,ldots,n
$$ for $q<infty$, and $|x|_infty le 1$ for $q=infty$.
Summing over $i$ yields
$$
|x|_q^q =|x_1|^q+|x_2|^q+cdots +|x_n|^qle |x_1|^p+|x_2|^p+cdots +|x_n|^p=1
$$ for $q<infty$. This proves $|x|_qle 1$ for all $1le p< qle infty$ as desired.
Note: The inequality can be easily extended to $Bbb R^{Bbb N}$ equipped with $l^p$-norms, i.e. $l^p(Bbb N)$. And reverse inequality is true if we are on a finite measure space, i.e.
$$
|f|_ple mu(Omega)^{1/p-1/q}|f|_q,quad 1le p<qleinfty.
$$ holds for $fin L^q(Omega)subseteq L^p(Omega)$. (It a direct consequence of Jensen's inequality.) But in a general measure space, in particular with infinite total measure such as $L^p(Bbb R)$, none of the relationship $$
|f|_ple C|f|_q
$$ nor
$$
|f|_qle C|f|_p
$$ is true for any $C>0$. (So there is no inclusion between $L^p(Bbb R)$-spaces. I'll omit further explanation, but one maybe able to find examples of this.) Thus the fact that $$l^p(Bbb N)subseteq l^q(Bbb N), quad 1le p<qle infty$$ is a special property, which can be attributed to certain properties of counting measure.
answered Jan 23 at 18:52
SongSong
17.5k21246
$endgroup$
$begingroup$
Are you sure? In my lecture it s written : $$ abs(f_m(t) - f_n(t)) leq vert vert f_m- f_n vert vert_{infty} $$ and also for any element in the space of sequences $l^2_{ mathbb F} $ : $$ abs( u_j^{(n) } - u_j^{(m) }) leq vert vert u^{(n) } - u^{(m) } vert vert_{2} $$
$endgroup$
– Marine Galantin
Jan 23 at 19:20
$begingroup$
And what about the other points I was talking about? Maybe it s linked to my other comment. Please be as precise as you can, i m baffled with all of this
$endgroup$
– Marine Galantin
Jan 23 at 19:24
1
$begingroup$
@MarineGalantin Yes, I'm so sure. If you are suspicious, why don't you calculate actual $p$-norms of a vector, say $(1,3,4)$? I think I fully explained why $|x|_p ge |x|_q$ is true. And about those two facts, yes, they are true $textbf{independently}$ of the inequality I showed. Those are just trivial pointwise bound obtained from the definition of sup-norm and $2$-norm. And they have nothing to do with the inequality discussed.
$endgroup$
– Song
Jan 23 at 19:43
1
$begingroup$
@MarineGalantin The direction of the inequality is clear if you remember the obvious endpoint $$ |x|_{ell^infty} le |x|_{ell^1}$$ that cannot be reversed
$endgroup$
– Calvin Khor
Jan 23 at 20:36
1
$begingroup$
@MarineGalantin Yes, surely! To add some more explanation (using symbolic manipulations), let us denote $ncdot B_p ={ncdot xin l^p(Bbb N);:;x in B_p}$ (it is just a ball of radius $n$.) Then,$$ forall nge 1,quad ncdot B_psubset ncdot B_qsubset l^q(Bbb N).$$ Since $bigcup_{nge 1} ncdot B_p = l^p(Bbb N)$ , it follows $$l^p(Bbb N)subset l^q(Bbb N).$$
$endgroup$
– Song
Jan 23 at 21:02
|
show 1 more comment
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$begingroup$
Actually, direction should be reversed: it holds that
$$
|x|_pge |x|_q,quad 1le p< qleinfty.tag{*}
$$ What your figure shows us is that
$$
B_psubset B_q,quad p< qtag{**}
$$ where $B_r$ is a unit ball in $r$-norm. It is easy to see that $text{(*)}$ is equivalent to $text{(**)}$.
Now, we prove $text{(*)}$. By the homogeneity of norms, it is sufficient to show that
$$
|x|_qle 1
$$ provided that $|x|_p= 1$. By the assumption, we have
$$
|x_i|^ple |x_1|^p+|x_2|^p+cdots +|x_n|^p=1
$$ for all $ile n$. Hence $|x_i|le 1$ and it follows
$$
|x_i|^qle |x_i|^p,quad i=1,2,ldots,n
$$ for $q<infty$, and $|x|_infty le 1$ for $q=infty$.
Summing over $i$ yields
$$
|x|_q^q =|x_1|^q+|x_2|^q+cdots +|x_n|^qle |x_1|^p+|x_2|^p+cdots +|x_n|^p=1
$$ for $q<infty$. This proves $|x|_qle 1$ for all $1le p< qle infty$ as desired.
Note: The inequality can be easily extended to $Bbb R^{Bbb N}$ equipped with $l^p$-norms, i.e. $l^p(Bbb N)$. And reverse inequality is true if we are on a finite measure space, i.e.
$$
|f|_ple mu(Omega)^{1/p-1/q}|f|_q,quad 1le p<qleinfty.
$$ holds for $fin L^q(Omega)subseteq L^p(Omega)$. (It a direct consequence of Jensen's inequality.) But in a general measure space, in particular with infinite total measure such as $L^p(Bbb R)$, none of the relationship $$
|f|_ple C|f|_q
$$ nor
$$
|f|_qle C|f|_p
$$ is true for any $C>0$. (So there is no inclusion between $L^p(Bbb R)$-spaces. I'll omit further explanation, but one maybe able to find examples of this.) Thus the fact that $$l^p(Bbb N)subseteq l^q(Bbb N), quad 1le p<qle infty$$ is a special property, which can be attributed to certain properties of counting measure.
answered Jan 23 at 18:52
SongSong
17.5k21246
$endgroup$
$begingroup$
Are you sure? In my lecture it s written : $$ abs(f_m(t) - f_n(t)) leq vert vert f_m- f_n vert vert_{infty} $$ and also for any element in the space of sequences $l^2_{ mathbb F} $ : $$ abs( u_j^{(n) } - u_j^{(m) }) leq vert vert u^{(n) } - u^{(m) } vert vert_{2} $$
$endgroup$
– Marine Galantin
Jan 23 at 19:20
$begingroup$
And what about the other points I was talking about? Maybe it s linked to my other comment. Please be as precise as you can, i m baffled with all of this
$endgroup$
– Marine Galantin
Jan 23 at 19:24
1
$begingroup$
@MarineGalantin Yes, I'm so sure. If you are suspicious, why don't you calculate actual $p$-norms of a vector, say $(1,3,4)$? I think I fully explained why $|x|_p ge |x|_q$ is true. And about those two facts, yes, they are true $textbf{independently}$ of the inequality I showed. Those are just trivial pointwise bound obtained from the definition of sup-norm and $2$-norm. And they have nothing to do with the inequality discussed.
$endgroup$
– Song
Jan 23 at 19:43
1
$begingroup$
@MarineGalantin The direction of the inequality is clear if you remember the obvious endpoint $$ |x|_{ell^infty} le |x|_{ell^1}$$ that cannot be reversed
$endgroup$
– Calvin Khor
Jan 23 at 20:36
1
$begingroup$
@MarineGalantin Yes, surely! To add some more explanation (using symbolic manipulations), let us denote $ncdot B_p ={ncdot xin l^p(Bbb N);:;x in B_p}$ (it is just a ball of radius $n$.) Then,$$ forall nge 1,quad ncdot B_psubset ncdot B_qsubset l^q(Bbb N).$$ Since $bigcup_{nge 1} ncdot B_p = l^p(Bbb N)$ , it follows $$l^p(Bbb N)subset l^q(Bbb N).$$
$endgroup$
– Song
Jan 23 at 21:02
|
show 1 more comment
$begingroup$
Actually, direction should be reversed: it holds that
$$
|x|_pge |x|_q,quad 1le p< qleinfty.tag{*}
$$ What your figure shows us is that
$$
B_psubset B_q,quad p< qtag{**}
$$ where $B_r$ is a unit ball in $r$-norm. It is easy to see that $text{(*)}$ is equivalent to $text{(**)}$.
Now, we prove $text{(*)}$. By the homogeneity of norms, it is sufficient to show that
$$
|x|_qle 1
$$ provided that $|x|_p= 1$. By the assumption, we have
$$
|x_i|^ple |x_1|^p+|x_2|^p+cdots +|x_n|^p=1
$$ for all $ile n$. Hence $|x_i|le 1$ and it follows
$$
|x_i|^qle |x_i|^p,quad i=1,2,ldots,n
$$ for $q<infty$, and $|x|_infty le 1$ for $q=infty$.
Summing over $i$ yields
$$
|x|_q^q =|x_1|^q+|x_2|^q+cdots +|x_n|^qle |x_1|^p+|x_2|^p+cdots +|x_n|^p=1
$$ for $q<infty$. This proves $|x|_qle 1$ for all $1le p< qle infty$ as desired.
Note: The inequality can be easily extended to $Bbb R^{Bbb N}$ equipped with $l^p$-norms, i.e. $l^p(Bbb N)$. And reverse inequality is true if we are on a finite measure space, i.e.
$$
|f|_ple mu(Omega)^{1/p-1/q}|f|_q,quad 1le p<qleinfty.
$$ holds for $fin L^q(Omega)subseteq L^p(Omega)$. (It a direct consequence of Jensen's inequality.) But in a general measure space, in particular with infinite total measure such as $L^p(Bbb R)$, none of the relationship $$
|f|_ple C|f|_q
$$ nor
$$
|f|_qle C|f|_p
$$ is true for any $C>0$. (So there is no inclusion between $L^p(Bbb R)$-spaces. I'll omit further explanation, but one maybe able to find examples of this.) Thus the fact that $$l^p(Bbb N)subseteq l^q(Bbb N), quad 1le p<qle infty$$ is a special property, which can be attributed to certain properties of counting measure.
answered Jan 23 at 18:52
SongSong
17.5k21246
$endgroup$
$begingroup$
Are you sure? In my lecture it s written : $$ abs(f_m(t) - f_n(t)) leq vert vert f_m- f_n vert vert_{infty} $$ and also for any element in the space of sequences $l^2_{ mathbb F} $ : $$ abs( u_j^{(n) } - u_j^{(m) }) leq vert vert u^{(n) } - u^{(m) } vert vert_{2} $$
$endgroup$
– Marine Galantin
Jan 23 at 19:20
$begingroup$
And what about the other points I was talking about? Maybe it s linked to my other comment. Please be as precise as you can, i m baffled with all of this
$endgroup$
– Marine Galantin
Jan 23 at 19:24
1
$begingroup$
@MarineGalantin Yes, I'm so sure. If you are suspicious, why don't you calculate actual $p$-norms of a vector, say $(1,3,4)$? I think I fully explained why $|x|_p ge |x|_q$ is true. And about those two facts, yes, they are true $textbf{independently}$ of the inequality I showed. Those are just trivial pointwise bound obtained from the definition of sup-norm and $2$-norm. And they have nothing to do with the inequality discussed.
$endgroup$
– Song
Jan 23 at 19:43
1
$begingroup$
@MarineGalantin The direction of the inequality is clear if you remember the obvious endpoint $$ |x|_{ell^infty} le |x|_{ell^1}$$ that cannot be reversed
$endgroup$
– Calvin Khor
Jan 23 at 20:36
1
$begingroup$
@MarineGalantin Yes, surely! To add some more explanation (using symbolic manipulations), let us denote $ncdot B_p ={ncdot xin l^p(Bbb N);:;x in B_p}$ (it is just a ball of radius $n$.) Then,$$ forall nge 1,quad ncdot B_psubset ncdot B_qsubset l^q(Bbb N).$$ Since $bigcup_{nge 1} ncdot B_p = l^p(Bbb N)$ , it follows $$l^p(Bbb N)subset l^q(Bbb N).$$
$endgroup$
– Song
Jan 23 at 21:02
|
show 1 more comment
$begingroup$
Actually, direction should be reversed: it holds that
$$
|x|_pge |x|_q,quad 1le p< qleinfty.tag{*}
$$ What your figure shows us is that
$$
B_psubset B_q,quad p< qtag{**}
$$ where $B_r$ is a unit ball in $r$-norm. It is easy to see that $text{(*)}$ is equivalent to $text{(**)}$.
Now, we prove $text{(*)}$. By the homogeneity of norms, it is sufficient to show that
$$
|x|_qle 1
$$ provided that $|x|_p= 1$. By the assumption, we have
$$
|x_i|^ple |x_1|^p+|x_2|^p+cdots +|x_n|^p=1
$$ for all $ile n$. Hence $|x_i|le 1$ and it follows
$$
|x_i|^qle |x_i|^p,quad i=1,2,ldots,n
$$ for $q<infty$, and $|x|_infty le 1$ for $q=infty$.
Summing over $i$ yields
$$
|x|_q^q =|x_1|^q+|x_2|^q+cdots +|x_n|^qle |x_1|^p+|x_2|^p+cdots +|x_n|^p=1
$$ for $q<infty$. This proves $|x|_qle 1$ for all $1le p< qle infty$ as desired.
Note: The inequality can be easily extended to $Bbb R^{Bbb N}$ equipped with $l^p$-norms, i.e. $l^p(Bbb N)$. And reverse inequality is true if we are on a finite measure space, i.e.
$$
|f|_ple mu(Omega)^{1/p-1/q}|f|_q,quad 1le p<qleinfty.
$$ holds for $fin L^q(Omega)subseteq L^p(Omega)$. (It a direct consequence of Jensen's inequality.) But in a general measure space, in particular with infinite total measure such as $L^p(Bbb R)$, none of the relationship $$
|f|_ple C|f|_q
$$ nor
$$
|f|_qle C|f|_p
$$ is true for any $C>0$. (So there is no inclusion between $L^p(Bbb R)$-spaces. I'll omit further explanation, but one maybe able to find examples of this.) Thus the fact that $$l^p(Bbb N)subseteq l^q(Bbb N), quad 1le p<qle infty$$ is a special property, which can be attributed to certain properties of counting measure.
answered Jan 23 at 18:52
SongSong
17.5k21246
$endgroup$
Actually, direction should be reversed: it holds that
$$
|x|_pge |x|_q,quad 1le p< qleinfty.tag{*}
$$ What your figure shows us is that
$$
B_psubset B_q,quad p< qtag{**}
$$ where $B_r$ is a unit ball in $r$-norm. It is easy to see that $text{(*)}$ is equivalent to $text{(**)}$.
Now, we prove $text{(*)}$. By the homogeneity of norms, it is sufficient to show that
$$
|x|_qle 1
$$ provided that $|x|_p= 1$. By the assumption, we have
$$
|x_i|^ple |x_1|^p+|x_2|^p+cdots +|x_n|^p=1
$$ for all $ile n$. Hence $|x_i|le 1$ and it follows
$$
|x_i|^qle |x_i|^p,quad i=1,2,ldots,n
$$ for $q<infty$, and $|x|_infty le 1$ for $q=infty$.
Summing over $i$ yields
$$
|x|_q^q =|x_1|^q+|x_2|^q+cdots +|x_n|^qle |x_1|^p+|x_2|^p+cdots +|x_n|^p=1
$$ for $q<infty$. This proves $|x|_qle 1$ for all $1le p< qle infty$ as desired.
Note: The inequality can be easily extended to $Bbb R^{Bbb N}$ equipped with $l^p$-norms, i.e. $l^p(Bbb N)$. And reverse inequality is true if we are on a finite measure space, i.e.
$$
|f|_ple mu(Omega)^{1/p-1/q}|f|_q,quad 1le p<qleinfty.
$$ holds for $fin L^q(Omega)subseteq L^p(Omega)$. (It a direct consequence of Jensen's inequality.) But in a general measure space, in particular with infinite total measure such as $L^p(Bbb R)$, none of the relationship $$
|f|_ple C|f|_q
$$ nor
$$
|f|_qle C|f|_p
$$ is true for any $C>0$. (So there is no inclusion between $L^p(Bbb R)$-spaces. I'll omit further explanation, but one maybe able to find examples of this.) Thus the fact that $$l^p(Bbb N)subseteq l^q(Bbb N), quad 1le p<qle infty$$ is a special property, which can be attributed to certain properties of counting measure.
answered Jan 23 at 18:52
SongSong
17.5k21246
edited Jan 23 at 19:56
answered Jan 23 at 18:52
SongSong
17.5k21246
answered Jan 23 at 18:52
SongSong
17.5k21246
answered Jan 23 at 18:52
SongSong
17.5k21246
17.5k21246
$begingroup$
Are you sure? In my lecture it s written : $$ abs(f_m(t) - f_n(t)) leq vert vert f_m- f_n vert vert_{infty} $$ and also for any element in the space of sequences $l^2_{ mathbb F} $ : $$ abs( u_j^{(n) } - u_j^{(m) }) leq vert vert u^{(n) } - u^{(m) } vert vert_{2} $$
$endgroup$
– Marine Galantin
Jan 23 at 19:20
$begingroup$
And what about the other points I was talking about? Maybe it s linked to my other comment. Please be as precise as you can, i m baffled with all of this
$endgroup$
– Marine Galantin
Jan 23 at 19:24
1
$begingroup$
@MarineGalantin Yes, I'm so sure. If you are suspicious, why don't you calculate actual $p$-norms of a vector, say $(1,3,4)$? I think I fully explained why $|x|_p ge |x|_q$ is true. And about those two facts, yes, they are true $textbf{independently}$ of the inequality I showed. Those are just trivial pointwise bound obtained from the definition of sup-norm and $2$-norm. And they have nothing to do with the inequality discussed.
$endgroup$
– Song
Jan 23 at 19:43
1
$begingroup$
@MarineGalantin The direction of the inequality is clear if you remember the obvious endpoint $$ |x|_{ell^infty} le |x|_{ell^1}$$ that cannot be reversed
$endgroup$
– Calvin Khor
Jan 23 at 20:36
1
$begingroup$
@MarineGalantin Yes, surely! To add some more explanation (using symbolic manipulations), let us denote $ncdot B_p ={ncdot xin l^p(Bbb N);:;x in B_p}$ (it is just a ball of radius $n$.) Then,$$ forall nge 1,quad ncdot B_psubset ncdot B_qsubset l^q(Bbb N).$$ Since $bigcup_{nge 1} ncdot B_p = l^p(Bbb N)$ , it follows $$l^p(Bbb N)subset l^q(Bbb N).$$
$endgroup$
– Song
Jan 23 at 21:02
|
show 1 more comment
$begingroup$
Are you sure? In my lecture it s written : $$ abs(f_m(t) - f_n(t)) leq vert vert f_m- f_n vert vert_{infty} $$ and also for any element in the space of sequences $l^2_{ mathbb F} $ : $$ abs( u_j^{(n) } - u_j^{(m) }) leq vert vert u^{(n) } - u^{(m) } vert vert_{2} $$
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– Marine Galantin
Jan 23 at 19:20
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And what about the other points I was talking about? Maybe it s linked to my other comment. Please be as precise as you can, i m baffled with all of this
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– Marine Galantin
Jan 23 at 19:24
1
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@MarineGalantin Yes, I'm so sure. If you are suspicious, why don't you calculate actual $p$-norms of a vector, say $(1,3,4)$? I think I fully explained why $|x|_p ge |x|_q$ is true. And about those two facts, yes, they are true $textbf{independently}$ of the inequality I showed. Those are just trivial pointwise bound obtained from the definition of sup-norm and $2$-norm. And they have nothing to do with the inequality discussed.
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– Song
Jan 23 at 19:43
1
$begingroup$
@MarineGalantin The direction of the inequality is clear if you remember the obvious endpoint $$ |x|_{ell^infty} le |x|_{ell^1}$$ that cannot be reversed
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– Calvin Khor
Jan 23 at 20:36
1
$begingroup$
@MarineGalantin Yes, surely! To add some more explanation (using symbolic manipulations), let us denote $ncdot B_p ={ncdot xin l^p(Bbb N);:;x in B_p}$ (it is just a ball of radius $n$.) Then,$$ forall nge 1,quad ncdot B_psubset ncdot B_qsubset l^q(Bbb N).$$ Since $bigcup_{nge 1} ncdot B_p = l^p(Bbb N)$ , it follows $$l^p(Bbb N)subset l^q(Bbb N).$$
$endgroup$
– Song
Jan 23 at 21:02
$begingroup$
Are you sure? In my lecture it s written : $$ abs(f_m(t) - f_n(t)) leq vert vert f_m- f_n vert vert_{infty} $$ and also for any element in the space of sequences $l^2_{ mathbb F} $ : $$ abs( u_j^{(n) } - u_j^{(m) }) leq vert vert u^{(n) } - u^{(m) } vert vert_{2} $$
$endgroup$
– Marine Galantin
Jan 23 at 19:20
$begingroup$
Are you sure? In my lecture it s written : $$ abs(f_m(t) - f_n(t)) leq vert vert f_m- f_n vert vert_{infty} $$ and also for any element in the space of sequences $l^2_{ mathbb F} $ : $$ abs( u_j^{(n) } - u_j^{(m) }) leq vert vert u^{(n) } - u^{(m) } vert vert_{2} $$
$endgroup$
– Marine Galantin
Jan 23 at 19:20
$begingroup$
And what about the other points I was talking about? Maybe it s linked to my other comment. Please be as precise as you can, i m baffled with all of this
$endgroup$
– Marine Galantin
Jan 23 at 19:24
$begingroup$
And what about the other points I was talking about? Maybe it s linked to my other comment. Please be as precise as you can, i m baffled with all of this
$endgroup$
– Marine Galantin
Jan 23 at 19:24
1
1
$begingroup$
@MarineGalantin Yes, I'm so sure. If you are suspicious, why don't you calculate actual $p$-norms of a vector, say $(1,3,4)$? I think I fully explained why $|x|_p ge |x|_q$ is true. And about those two facts, yes, they are true $textbf{independently}$ of the inequality I showed. Those are just trivial pointwise bound obtained from the definition of sup-norm and $2$-norm. And they have nothing to do with the inequality discussed.
$endgroup$
– Song
Jan 23 at 19:43
$begingroup$
@MarineGalantin Yes, I'm so sure. If you are suspicious, why don't you calculate actual $p$-norms of a vector, say $(1,3,4)$? I think I fully explained why $|x|_p ge |x|_q$ is true. And about those two facts, yes, they are true $textbf{independently}$ of the inequality I showed. Those are just trivial pointwise bound obtained from the definition of sup-norm and $2$-norm. And they have nothing to do with the inequality discussed.
$endgroup$
– Song
Jan 23 at 19:43
1
1
$begingroup$
@MarineGalantin The direction of the inequality is clear if you remember the obvious endpoint $$ |x|_{ell^infty} le |x|_{ell^1}$$ that cannot be reversed
$endgroup$
– Calvin Khor
Jan 23 at 20:36
$begingroup$
@MarineGalantin The direction of the inequality is clear if you remember the obvious endpoint $$ |x|_{ell^infty} le |x|_{ell^1}$$ that cannot be reversed
$endgroup$
– Calvin Khor
Jan 23 at 20:36
1
1
$begingroup$
@MarineGalantin Yes, surely! To add some more explanation (using symbolic manipulations), let us denote $ncdot B_p ={ncdot xin l^p(Bbb N);:;x in B_p}$ (it is just a ball of radius $n$.) Then,$$ forall nge 1,quad ncdot B_psubset ncdot B_qsubset l^q(Bbb N).$$ Since $bigcup_{nge 1} ncdot B_p = l^p(Bbb N)$ , it follows $$l^p(Bbb N)subset l^q(Bbb N).$$
$endgroup$
– Song
Jan 23 at 21:02
$begingroup$
@MarineGalantin Yes, surely! To add some more explanation (using symbolic manipulations), let us denote $ncdot B_p ={ncdot xin l^p(Bbb N);:;x in B_p}$ (it is just a ball of radius $n$.) Then,$$ forall nge 1,quad ncdot B_psubset ncdot B_qsubset l^q(Bbb N).$$ Since $bigcup_{nge 1} ncdot B_p = l^p(Bbb N)$ , it follows $$l^p(Bbb N)subset l^q(Bbb N).$$
$endgroup$
– Song
Jan 23 at 21:02
|
show 1 more comment
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