Mixing
Finite dimensional normed vector space, projection continuous?
$begingroup$
Let $X$ be an $n$ -dimensional normed $mathbb{R}$-vector space. Let ${e_i mid i=1,...,n}$ be a basis of $X$. Is it true that the function $f:Xrightarrow mathbb{R}$, $sum lambda_i e_i mapsto lambda_i$ is continuous? I thought it would be easy to prove this but I wasn't able to.
Could you help me? Thanks!
linear-algebra general-topology analysis
edited Dec 31 '18 at 10:01
dmtri
1,4521521
asked Dec 31 '18 at 9:06
JiuJiu
471112
$endgroup$
add a comment |
$begingroup$
Let $X$ be an $n$ -dimensional normed $mathbb{R}$-vector space. Let ${e_i mid i=1,...,n}$ be a basis of $X$. Is it true that the function $f:Xrightarrow mathbb{R}$, $sum lambda_i e_i mapsto lambda_i$ is continuous? I thought it would be easy to prove this but I wasn't able to.
Could you help me? Thanks!
linear-algebra general-topology analysis
edited Dec 31 '18 at 10:01
dmtri
1,4521521
asked Dec 31 '18 at 9:06
JiuJiu
471112
$endgroup$
$begingroup$
The norm in $X$ induces a topology, namely open sets.
$endgroup$
– dmtri
Dec 31 '18 at 9:36
$begingroup$
There are several theorems about finite dimensional normed linear spaces: any two norms are equivalent. any linear map on then is continuous etc. [These two theorems are easily seen to be equivalent]. They provide an answer to your question]. But if you want to avoid these theorems and use only the definition of norm you can look at my answer. @Jiu
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:08
$begingroup$
@KaviRamaMurthy In fact I was trying to prove that any two norms on a finite dimensional space are equivalent. And I needed the continuity of $f$ for my proof.
$endgroup$
– Jiu
Dec 31 '18 at 10:11
add a comment |
$begingroup$
Let $X$ be an $n$ -dimensional normed $mathbb{R}$-vector space. Let ${e_i mid i=1,...,n}$ be a basis of $X$. Is it true that the function $f:Xrightarrow mathbb{R}$, $sum lambda_i e_i mapsto lambda_i$ is continuous? I thought it would be easy to prove this but I wasn't able to.
Could you help me? Thanks!
linear-algebra general-topology analysis
edited Dec 31 '18 at 10:01
dmtri
1,4521521
asked Dec 31 '18 at 9:06
JiuJiu
471112
$endgroup$
Let $X$ be an $n$ -dimensional normed $mathbb{R}$-vector space. Let ${e_i mid i=1,...,n}$ be a basis of $X$. Is it true that the function $f:Xrightarrow mathbb{R}$, $sum lambda_i e_i mapsto lambda_i$ is continuous? I thought it would be easy to prove this but I wasn't able to.
Could you help me? Thanks!
linear-algebra general-topology analysis
linear-algebra general-topology analysis
edited Dec 31 '18 at 10:01
dmtri
1,4521521
asked Dec 31 '18 at 9:06
JiuJiu
471112
edited Dec 31 '18 at 10:01
dmtri
1,4521521
asked Dec 31 '18 at 9:06
JiuJiu
471112
edited Dec 31 '18 at 10:01
dmtri
1,4521521
edited Dec 31 '18 at 10:01
dmtri
1,4521521
edited Dec 31 '18 at 10:01
dmtri
1,4521521
1,4521521
asked Dec 31 '18 at 9:06
JiuJiu
471112
asked Dec 31 '18 at 9:06
JiuJiu
471112
asked Dec 31 '18 at 9:06
JiuJiu
471112
471112
$begingroup$
The norm in $X$ induces a topology, namely open sets.
$endgroup$
– dmtri
Dec 31 '18 at 9:36
$begingroup$
There are several theorems about finite dimensional normed linear spaces: any two norms are equivalent. any linear map on then is continuous etc. [These two theorems are easily seen to be equivalent]. They provide an answer to your question]. But if you want to avoid these theorems and use only the definition of norm you can look at my answer. @Jiu
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:08
$begingroup$
@KaviRamaMurthy In fact I was trying to prove that any two norms on a finite dimensional space are equivalent. And I needed the continuity of $f$ for my proof.
$endgroup$
– Jiu
Dec 31 '18 at 10:11
add a comment |
$begingroup$
The norm in $X$ induces a topology, namely open sets.
$endgroup$
– dmtri
Dec 31 '18 at 9:36
$begingroup$
There are several theorems about finite dimensional normed linear spaces: any two norms are equivalent. any linear map on then is continuous etc. [These two theorems are easily seen to be equivalent]. They provide an answer to your question]. But if you want to avoid these theorems and use only the definition of norm you can look at my answer. @Jiu
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:08
$begingroup$
@KaviRamaMurthy In fact I was trying to prove that any two norms on a finite dimensional space are equivalent. And I needed the continuity of $f$ for my proof.
$endgroup$
– Jiu
Dec 31 '18 at 10:11
$begingroup$
The norm in $X$ induces a topology, namely open sets.
$endgroup$
– dmtri
Dec 31 '18 at 9:36
$begingroup$
The norm in $X$ induces a topology, namely open sets.
$endgroup$
– dmtri
Dec 31 '18 at 9:36
$begingroup$
There are several theorems about finite dimensional normed linear spaces: any two norms are equivalent. any linear map on then is continuous etc. [These two theorems are easily seen to be equivalent]. They provide an answer to your question]. But if you want to avoid these theorems and use only the definition of norm you can look at my answer. @Jiu
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:08
$begingroup$
There are several theorems about finite dimensional normed linear spaces: any two norms are equivalent. any linear map on then is continuous etc. [These two theorems are easily seen to be equivalent]. They provide an answer to your question]. But if you want to avoid these theorems and use only the definition of norm you can look at my answer. @Jiu
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:08
$begingroup$
@KaviRamaMurthy In fact I was trying to prove that any two norms on a finite dimensional space are equivalent. And I needed the continuity of $f$ for my proof.
$endgroup$
– Jiu
Dec 31 '18 at 10:11
$begingroup$
@KaviRamaMurthy In fact I was trying to prove that any two norms on a finite dimensional space are equivalent. And I needed the continuity of $f$ for my proof.
$endgroup$
– Jiu
Dec 31 '18 at 10:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here is a proof that does not use any theorem on general normed linear spaces. It uses only the Bolzano Weierstras Theorem for the usual norm on $mathbb R^{n}$. Suppose $sum a_i^{(m)}e_i to 0$. We have to show that $a_i^{(m)} to 0$ for each $i$. We claim that ${(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is bounded. Suppose, if possible, $leftVert
(a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})rightVert_2 to
infty $. Here $|.|_2$ denotes the usual norm on $mathbb R^{n}$. Let $b_{i}^{(m_{j})}=frac{a_{i}^{(m_{j})}}{leftVert
(a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})rightVert_2 },1leq
ileq n.$ Then $sum_{i=1}^{n}b_{i}^{(m)}x_{i}to 0$ and $%
leftVert (b_{1}^{(m_{j})},b_{2}^{(m_{j})},...,b_{n}^{(m_{j})})rightVert_2
=1$ $forall j$. Since unit vectors form a compact set in $mathbb R^{n}$ we can take limit of $sum_{i=1}^{n}b_{i}^{(m)}x_{i}$ through
a subsequence to get an equation of the type $
sum_{i=1}^{n}c_{i}x_{i}=0$ where $leftVert
(c_{1},c_{2},...,c_{n})rightVert_2 =1.$ This is impossible because $%
{x_{1},x_{2},...,x_{n}}$ be a basis. We have proved that $%
{(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is bounded. We now extract a
convergent subsequence of this sequence to get $
sum_{i=1}^{n}d_{i}x_{i}=0$ forcing each $d_{i}$ to be $0$. This shows
that every limit point of ${(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is
$(0,0,,,,0)$ and hence the claim is true.
answered Dec 31 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
$endgroup$
add a comment |
$begingroup$
Since any two norms on a finite dimensional space are equivalent, and $|sum_i lambda_i e_i|_* = sum_i |lambda_i|$ defines a norm, there exists a constant $C>0$ such that
$$
|sum_i lambda_i e_i|_*leq C|sum_i lambda_i e_i|,quadforall (lambda_i)_{i=1}^n.
$$It follows
$$
left|f(sum_i lambda_i e_i)right |= |lambda_j|le |sum_i lambda_i e_i|_*leq C|sum_i lambda_i e_i|.$$
EDIT: Here's a sketch of the proof that any two norms on a f.d.v.s. $V$ on $mathbb{F}$($=mathbb{R}$ or $mathbb{C}$) are equivalent. Let $(e_i)_{i=1}^n$ be a basis of $V$, and endow $V$ a norm defined by $|sum_i x_i e_i|_1 =sum_i |x_i|$. Then, $(V,|cdot|_1)$ and $(mathbb{F}^n,|cdot|_1)$ are isometric and hence homeomorphic. Notice that this implies ${sum_i x_ie_i;|;sum_i |x_i| = 1}$ is compact in $V$.
Now, let $f(sum_i x_ie_i) =|sum_i x_i e_i|$ where $|cdot|$ denotes the given norm of $V$. If we show $|cdot| sim|cdot|_1$, then the claim follows since $sim$ is an equivalence relation. Observe that
$$
|f(sum_i x_ie_i)-f(sum_i y_ie_i)| = ||sum_i x_ie_i|-|sum_i y_ie_i||leq |sum_i (x_i-y_i)e_i| leq sum_i |x_i-y_i|cdot max_i |e_i|.
$$ This shows $f:(V,|cdot|_1)to [0,infty)$ is continuous. Since $E={sum_i x_ie_i;|;sum_i |x_i| = 1}$ is compact, by extremum value theorem, there is $0<mleq M<infty$ such that
$$
mleq f(x) =|x|le M, quad forall xin E.
$$ This gives $m |x|_1 le |x|le M|x|_1$ for all $xin V$.
answered Dec 31 '18 at 9:29
SongSong
7,163421
$endgroup$
$begingroup$
Is there a way to prove that two norms on a finite dimensional space are equivalent without using the continuity of $f$? In other words, I want to make sure that it is not circular reasoning.
$endgroup$
– Jiu
Dec 31 '18 at 10:07
$begingroup$
Yes, the proof I know uses only topological tools.
$endgroup$
– Song
Dec 31 '18 at 10:21
add a comment |
$begingroup$
Hint: Let $P_1 v = v_1$. Observe
begin{align}
|v_1-w_1| = |P_1v-P_1w| leq |P_1||v-w|.
end{align}
Edit: A upon the request of @KaviRamaMurthy, I have decided to include a quick proof that all linear maps $A:mathbb{R}^n rightarrow mathbb{R}^n$ are bounded.
Using matrix representation, we see that
begin{align}
|Av|^2=& sum^n_{j=1}left|sum^n_{j=1} a_{ij}v_j right|^2 leq sum^n_{j=1}left( sum^n_{j=1} |a_{ij}|^2right)left( sum^n_{j=1} |v_j|^2right) \
=& left(sum^n_{i,j=1}|a_{ij}|^2 right)|v|^2
end{align}
which means $|Av| leq C|v|$ where $C$ is independent of $v$.
answered Dec 31 '18 at 9:18
Jacky ChongJacky Chong
17.8k21128
$endgroup$
$begingroup$
The norm given is an arbitrary norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 9:41
$begingroup$
This answer is wrong. It is valid for the usual norm on Euclidean spaces, not for a general norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:10
$begingroup$
@KaviRamaMurthy I think the answer is okay. Observe $|v_1-w_1| = |v_1 mathbf{e}_1 - w_1mathbf{e}_1| = |P_1 v-P_1w|leq |P_1|_text{op} |v-w|$
$endgroup$
– Jacky Chong
Dec 31 '18 at 23:46
$begingroup$
You have not understood the question. you are assuming what you are asked to prove. You are asked to show that $P_1$ is a continuous map. How can you assume that it is a bounded operator to prove that it is continuous? @Jacky Chong
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 23:58
$begingroup$
@KaviRamaMurthy I guess I'm assuming all linear maps on finite dimensional vector spaces are bounded.
$endgroup$
– Jacky Chong
Jan 1 at 0:03
|
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
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$begingroup$
Here is a proof that does not use any theorem on general normed linear spaces. It uses only the Bolzano Weierstras Theorem for the usual norm on $mathbb R^{n}$. Suppose $sum a_i^{(m)}e_i to 0$. We have to show that $a_i^{(m)} to 0$ for each $i$. We claim that ${(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is bounded. Suppose, if possible, $leftVert
(a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})rightVert_2 to
infty $. Here $|.|_2$ denotes the usual norm on $mathbb R^{n}$. Let $b_{i}^{(m_{j})}=frac{a_{i}^{(m_{j})}}{leftVert
(a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})rightVert_2 },1leq
ileq n.$ Then $sum_{i=1}^{n}b_{i}^{(m)}x_{i}to 0$ and $%
leftVert (b_{1}^{(m_{j})},b_{2}^{(m_{j})},...,b_{n}^{(m_{j})})rightVert_2
=1$ $forall j$. Since unit vectors form a compact set in $mathbb R^{n}$ we can take limit of $sum_{i=1}^{n}b_{i}^{(m)}x_{i}$ through
a subsequence to get an equation of the type $
sum_{i=1}^{n}c_{i}x_{i}=0$ where $leftVert
(c_{1},c_{2},...,c_{n})rightVert_2 =1.$ This is impossible because $%
{x_{1},x_{2},...,x_{n}}$ be a basis. We have proved that $%
{(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is bounded. We now extract a
convergent subsequence of this sequence to get $
sum_{i=1}^{n}d_{i}x_{i}=0$ forcing each $d_{i}$ to be $0$. This shows
that every limit point of ${(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is
$(0,0,,,,0)$ and hence the claim is true.
answered Dec 31 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
$endgroup$
add a comment |
$begingroup$
Here is a proof that does not use any theorem on general normed linear spaces. It uses only the Bolzano Weierstras Theorem for the usual norm on $mathbb R^{n}$. Suppose $sum a_i^{(m)}e_i to 0$. We have to show that $a_i^{(m)} to 0$ for each $i$. We claim that ${(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is bounded. Suppose, if possible, $leftVert
(a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})rightVert_2 to
infty $. Here $|.|_2$ denotes the usual norm on $mathbb R^{n}$. Let $b_{i}^{(m_{j})}=frac{a_{i}^{(m_{j})}}{leftVert
(a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})rightVert_2 },1leq
ileq n.$ Then $sum_{i=1}^{n}b_{i}^{(m)}x_{i}to 0$ and $%
leftVert (b_{1}^{(m_{j})},b_{2}^{(m_{j})},...,b_{n}^{(m_{j})})rightVert_2
=1$ $forall j$. Since unit vectors form a compact set in $mathbb R^{n}$ we can take limit of $sum_{i=1}^{n}b_{i}^{(m)}x_{i}$ through
a subsequence to get an equation of the type $
sum_{i=1}^{n}c_{i}x_{i}=0$ where $leftVert
(c_{1},c_{2},...,c_{n})rightVert_2 =1.$ This is impossible because $%
{x_{1},x_{2},...,x_{n}}$ be a basis. We have proved that $%
{(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is bounded. We now extract a
convergent subsequence of this sequence to get $
sum_{i=1}^{n}d_{i}x_{i}=0$ forcing each $d_{i}$ to be $0$. This shows
that every limit point of ${(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is
$(0,0,,,,0)$ and hence the claim is true.
answered Dec 31 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
$endgroup$
add a comment |
$begingroup$
Here is a proof that does not use any theorem on general normed linear spaces. It uses only the Bolzano Weierstras Theorem for the usual norm on $mathbb R^{n}$. Suppose $sum a_i^{(m)}e_i to 0$. We have to show that $a_i^{(m)} to 0$ for each $i$. We claim that ${(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is bounded. Suppose, if possible, $leftVert
(a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})rightVert_2 to
infty $. Here $|.|_2$ denotes the usual norm on $mathbb R^{n}$. Let $b_{i}^{(m_{j})}=frac{a_{i}^{(m_{j})}}{leftVert
(a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})rightVert_2 },1leq
ileq n.$ Then $sum_{i=1}^{n}b_{i}^{(m)}x_{i}to 0$ and $%
leftVert (b_{1}^{(m_{j})},b_{2}^{(m_{j})},...,b_{n}^{(m_{j})})rightVert_2
=1$ $forall j$. Since unit vectors form a compact set in $mathbb R^{n}$ we can take limit of $sum_{i=1}^{n}b_{i}^{(m)}x_{i}$ through
a subsequence to get an equation of the type $
sum_{i=1}^{n}c_{i}x_{i}=0$ where $leftVert
(c_{1},c_{2},...,c_{n})rightVert_2 =1.$ This is impossible because $%
{x_{1},x_{2},...,x_{n}}$ be a basis. We have proved that $%
{(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is bounded. We now extract a
convergent subsequence of this sequence to get $
sum_{i=1}^{n}d_{i}x_{i}=0$ forcing each $d_{i}$ to be $0$. This shows
that every limit point of ${(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is
$(0,0,,,,0)$ and hence the claim is true.
answered Dec 31 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
$endgroup$
Here is a proof that does not use any theorem on general normed linear spaces. It uses only the Bolzano Weierstras Theorem for the usual norm on $mathbb R^{n}$. Suppose $sum a_i^{(m)}e_i to 0$. We have to show that $a_i^{(m)} to 0$ for each $i$. We claim that ${(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is bounded. Suppose, if possible, $leftVert
(a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})rightVert_2 to
infty $. Here $|.|_2$ denotes the usual norm on $mathbb R^{n}$. Let $b_{i}^{(m_{j})}=frac{a_{i}^{(m_{j})}}{leftVert
(a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})rightVert_2 },1leq
ileq n.$ Then $sum_{i=1}^{n}b_{i}^{(m)}x_{i}to 0$ and $%
leftVert (b_{1}^{(m_{j})},b_{2}^{(m_{j})},...,b_{n}^{(m_{j})})rightVert_2
=1$ $forall j$. Since unit vectors form a compact set in $mathbb R^{n}$ we can take limit of $sum_{i=1}^{n}b_{i}^{(m)}x_{i}$ through
a subsequence to get an equation of the type $
sum_{i=1}^{n}c_{i}x_{i}=0$ where $leftVert
(c_{1},c_{2},...,c_{n})rightVert_2 =1.$ This is impossible because $%
{x_{1},x_{2},...,x_{n}}$ be a basis. We have proved that $%
{(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is bounded. We now extract a
convergent subsequence of this sequence to get $
sum_{i=1}^{n}d_{i}x_{i}=0$ forcing each $d_{i}$ to be $0$. This shows
that every limit point of ${(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})}$ is
$(0,0,,,,0)$ and hence the claim is true.
answered Dec 31 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
edited Dec 31 '18 at 9:46
answered Dec 31 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
answered Dec 31 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
answered Dec 31 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
52.5k32055
add a comment |
add a comment |
$begingroup$
Since any two norms on a finite dimensional space are equivalent, and $|sum_i lambda_i e_i|_* = sum_i |lambda_i|$ defines a norm, there exists a constant $C>0$ such that
$$
|sum_i lambda_i e_i|_*leq C|sum_i lambda_i e_i|,quadforall (lambda_i)_{i=1}^n.
$$It follows
$$
left|f(sum_i lambda_i e_i)right |= |lambda_j|le |sum_i lambda_i e_i|_*leq C|sum_i lambda_i e_i|.$$
EDIT: Here's a sketch of the proof that any two norms on a f.d.v.s. $V$ on $mathbb{F}$($=mathbb{R}$ or $mathbb{C}$) are equivalent. Let $(e_i)_{i=1}^n$ be a basis of $V$, and endow $V$ a norm defined by $|sum_i x_i e_i|_1 =sum_i |x_i|$. Then, $(V,|cdot|_1)$ and $(mathbb{F}^n,|cdot|_1)$ are isometric and hence homeomorphic. Notice that this implies ${sum_i x_ie_i;|;sum_i |x_i| = 1}$ is compact in $V$.
Now, let $f(sum_i x_ie_i) =|sum_i x_i e_i|$ where $|cdot|$ denotes the given norm of $V$. If we show $|cdot| sim|cdot|_1$, then the claim follows since $sim$ is an equivalence relation. Observe that
$$
|f(sum_i x_ie_i)-f(sum_i y_ie_i)| = ||sum_i x_ie_i|-|sum_i y_ie_i||leq |sum_i (x_i-y_i)e_i| leq sum_i |x_i-y_i|cdot max_i |e_i|.
$$ This shows $f:(V,|cdot|_1)to [0,infty)$ is continuous. Since $E={sum_i x_ie_i;|;sum_i |x_i| = 1}$ is compact, by extremum value theorem, there is $0<mleq M<infty$ such that
$$
mleq f(x) =|x|le M, quad forall xin E.
$$ This gives $m |x|_1 le |x|le M|x|_1$ for all $xin V$.
answered Dec 31 '18 at 9:29
SongSong
7,163421
$endgroup$
$begingroup$
Is there a way to prove that two norms on a finite dimensional space are equivalent without using the continuity of $f$? In other words, I want to make sure that it is not circular reasoning.
$endgroup$
– Jiu
Dec 31 '18 at 10:07
$begingroup$
Yes, the proof I know uses only topological tools.
$endgroup$
– Song
Dec 31 '18 at 10:21
add a comment |
$begingroup$
Since any two norms on a finite dimensional space are equivalent, and $|sum_i lambda_i e_i|_* = sum_i |lambda_i|$ defines a norm, there exists a constant $C>0$ such that
$$
|sum_i lambda_i e_i|_*leq C|sum_i lambda_i e_i|,quadforall (lambda_i)_{i=1}^n.
$$It follows
$$
left|f(sum_i lambda_i e_i)right |= |lambda_j|le |sum_i lambda_i e_i|_*leq C|sum_i lambda_i e_i|.$$
EDIT: Here's a sketch of the proof that any two norms on a f.d.v.s. $V$ on $mathbb{F}$($=mathbb{R}$ or $mathbb{C}$) are equivalent. Let $(e_i)_{i=1}^n$ be a basis of $V$, and endow $V$ a norm defined by $|sum_i x_i e_i|_1 =sum_i |x_i|$. Then, $(V,|cdot|_1)$ and $(mathbb{F}^n,|cdot|_1)$ are isometric and hence homeomorphic. Notice that this implies ${sum_i x_ie_i;|;sum_i |x_i| = 1}$ is compact in $V$.
Now, let $f(sum_i x_ie_i) =|sum_i x_i e_i|$ where $|cdot|$ denotes the given norm of $V$. If we show $|cdot| sim|cdot|_1$, then the claim follows since $sim$ is an equivalence relation. Observe that
$$
|f(sum_i x_ie_i)-f(sum_i y_ie_i)| = ||sum_i x_ie_i|-|sum_i y_ie_i||leq |sum_i (x_i-y_i)e_i| leq sum_i |x_i-y_i|cdot max_i |e_i|.
$$ This shows $f:(V,|cdot|_1)to [0,infty)$ is continuous. Since $E={sum_i x_ie_i;|;sum_i |x_i| = 1}$ is compact, by extremum value theorem, there is $0<mleq M<infty$ such that
$$
mleq f(x) =|x|le M, quad forall xin E.
$$ This gives $m |x|_1 le |x|le M|x|_1$ for all $xin V$.
answered Dec 31 '18 at 9:29
SongSong
7,163421
$endgroup$
$begingroup$
Is there a way to prove that two norms on a finite dimensional space are equivalent without using the continuity of $f$? In other words, I want to make sure that it is not circular reasoning.
$endgroup$
– Jiu
Dec 31 '18 at 10:07
$begingroup$
Yes, the proof I know uses only topological tools.
$endgroup$
– Song
Dec 31 '18 at 10:21
add a comment |
$begingroup$
Since any two norms on a finite dimensional space are equivalent, and $|sum_i lambda_i e_i|_* = sum_i |lambda_i|$ defines a norm, there exists a constant $C>0$ such that
$$
|sum_i lambda_i e_i|_*leq C|sum_i lambda_i e_i|,quadforall (lambda_i)_{i=1}^n.
$$It follows
$$
left|f(sum_i lambda_i e_i)right |= |lambda_j|le |sum_i lambda_i e_i|_*leq C|sum_i lambda_i e_i|.$$
EDIT: Here's a sketch of the proof that any two norms on a f.d.v.s. $V$ on $mathbb{F}$($=mathbb{R}$ or $mathbb{C}$) are equivalent. Let $(e_i)_{i=1}^n$ be a basis of $V$, and endow $V$ a norm defined by $|sum_i x_i e_i|_1 =sum_i |x_i|$. Then, $(V,|cdot|_1)$ and $(mathbb{F}^n,|cdot|_1)$ are isometric and hence homeomorphic. Notice that this implies ${sum_i x_ie_i;|;sum_i |x_i| = 1}$ is compact in $V$.
Now, let $f(sum_i x_ie_i) =|sum_i x_i e_i|$ where $|cdot|$ denotes the given norm of $V$. If we show $|cdot| sim|cdot|_1$, then the claim follows since $sim$ is an equivalence relation. Observe that
$$
|f(sum_i x_ie_i)-f(sum_i y_ie_i)| = ||sum_i x_ie_i|-|sum_i y_ie_i||leq |sum_i (x_i-y_i)e_i| leq sum_i |x_i-y_i|cdot max_i |e_i|.
$$ This shows $f:(V,|cdot|_1)to [0,infty)$ is continuous. Since $E={sum_i x_ie_i;|;sum_i |x_i| = 1}$ is compact, by extremum value theorem, there is $0<mleq M<infty$ such that
$$
mleq f(x) =|x|le M, quad forall xin E.
$$ This gives $m |x|_1 le |x|le M|x|_1$ for all $xin V$.
answered Dec 31 '18 at 9:29
SongSong
7,163421
$endgroup$
Since any two norms on a finite dimensional space are equivalent, and $|sum_i lambda_i e_i|_* = sum_i |lambda_i|$ defines a norm, there exists a constant $C>0$ such that
$$
|sum_i lambda_i e_i|_*leq C|sum_i lambda_i e_i|,quadforall (lambda_i)_{i=1}^n.
$$It follows
$$
left|f(sum_i lambda_i e_i)right |= |lambda_j|le |sum_i lambda_i e_i|_*leq C|sum_i lambda_i e_i|.$$
EDIT: Here's a sketch of the proof that any two norms on a f.d.v.s. $V$ on $mathbb{F}$($=mathbb{R}$ or $mathbb{C}$) are equivalent. Let $(e_i)_{i=1}^n$ be a basis of $V$, and endow $V$ a norm defined by $|sum_i x_i e_i|_1 =sum_i |x_i|$. Then, $(V,|cdot|_1)$ and $(mathbb{F}^n,|cdot|_1)$ are isometric and hence homeomorphic. Notice that this implies ${sum_i x_ie_i;|;sum_i |x_i| = 1}$ is compact in $V$.
Now, let $f(sum_i x_ie_i) =|sum_i x_i e_i|$ where $|cdot|$ denotes the given norm of $V$. If we show $|cdot| sim|cdot|_1$, then the claim follows since $sim$ is an equivalence relation. Observe that
$$
|f(sum_i x_ie_i)-f(sum_i y_ie_i)| = ||sum_i x_ie_i|-|sum_i y_ie_i||leq |sum_i (x_i-y_i)e_i| leq sum_i |x_i-y_i|cdot max_i |e_i|.
$$ This shows $f:(V,|cdot|_1)to [0,infty)$ is continuous. Since $E={sum_i x_ie_i;|;sum_i |x_i| = 1}$ is compact, by extremum value theorem, there is $0<mleq M<infty$ such that
$$
mleq f(x) =|x|le M, quad forall xin E.
$$ This gives $m |x|_1 le |x|le M|x|_1$ for all $xin V$.
answered Dec 31 '18 at 9:29
SongSong
7,163421
edited Dec 31 '18 at 10:33
answered Dec 31 '18 at 9:29
SongSong
7,163421
answered Dec 31 '18 at 9:29
SongSong
7,163421
answered Dec 31 '18 at 9:29
SongSong
7,163421
7,163421
$begingroup$
Is there a way to prove that two norms on a finite dimensional space are equivalent without using the continuity of $f$? In other words, I want to make sure that it is not circular reasoning.
$endgroup$
– Jiu
Dec 31 '18 at 10:07
$begingroup$
Yes, the proof I know uses only topological tools.
$endgroup$
– Song
Dec 31 '18 at 10:21
add a comment |
$begingroup$
Is there a way to prove that two norms on a finite dimensional space are equivalent without using the continuity of $f$? In other words, I want to make sure that it is not circular reasoning.
$endgroup$
– Jiu
Dec 31 '18 at 10:07
$begingroup$
Yes, the proof I know uses only topological tools.
$endgroup$
– Song
Dec 31 '18 at 10:21
$begingroup$
Is there a way to prove that two norms on a finite dimensional space are equivalent without using the continuity of $f$? In other words, I want to make sure that it is not circular reasoning.
$endgroup$
– Jiu
Dec 31 '18 at 10:07
$begingroup$
Is there a way to prove that two norms on a finite dimensional space are equivalent without using the continuity of $f$? In other words, I want to make sure that it is not circular reasoning.
$endgroup$
– Jiu
Dec 31 '18 at 10:07
$begingroup$
Yes, the proof I know uses only topological tools.
$endgroup$
– Song
Dec 31 '18 at 10:21
$begingroup$
Yes, the proof I know uses only topological tools.
$endgroup$
– Song
Dec 31 '18 at 10:21
add a comment |
$begingroup$
Hint: Let $P_1 v = v_1$. Observe
begin{align}
|v_1-w_1| = |P_1v-P_1w| leq |P_1||v-w|.
end{align}
Edit: A upon the request of @KaviRamaMurthy, I have decided to include a quick proof that all linear maps $A:mathbb{R}^n rightarrow mathbb{R}^n$ are bounded.
Using matrix representation, we see that
begin{align}
|Av|^2=& sum^n_{j=1}left|sum^n_{j=1} a_{ij}v_j right|^2 leq sum^n_{j=1}left( sum^n_{j=1} |a_{ij}|^2right)left( sum^n_{j=1} |v_j|^2right) \
=& left(sum^n_{i,j=1}|a_{ij}|^2 right)|v|^2
end{align}
which means $|Av| leq C|v|$ where $C$ is independent of $v$.
answered Dec 31 '18 at 9:18
Jacky ChongJacky Chong
17.8k21128
$endgroup$
$begingroup$
The norm given is an arbitrary norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 9:41
$begingroup$
This answer is wrong. It is valid for the usual norm on Euclidean spaces, not for a general norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:10
$begingroup$
@KaviRamaMurthy I think the answer is okay. Observe $|v_1-w_1| = |v_1 mathbf{e}_1 - w_1mathbf{e}_1| = |P_1 v-P_1w|leq |P_1|_text{op} |v-w|$
$endgroup$
– Jacky Chong
Dec 31 '18 at 23:46
$begingroup$
You have not understood the question. you are assuming what you are asked to prove. You are asked to show that $P_1$ is a continuous map. How can you assume that it is a bounded operator to prove that it is continuous? @Jacky Chong
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 23:58
$begingroup$
@KaviRamaMurthy I guess I'm assuming all linear maps on finite dimensional vector spaces are bounded.
$endgroup$
– Jacky Chong
Jan 1 at 0:03
|
show 5 more comments
$begingroup$
Hint: Let $P_1 v = v_1$. Observe
begin{align}
|v_1-w_1| = |P_1v-P_1w| leq |P_1||v-w|.
end{align}
Edit: A upon the request of @KaviRamaMurthy, I have decided to include a quick proof that all linear maps $A:mathbb{R}^n rightarrow mathbb{R}^n$ are bounded.
Using matrix representation, we see that
begin{align}
|Av|^2=& sum^n_{j=1}left|sum^n_{j=1} a_{ij}v_j right|^2 leq sum^n_{j=1}left( sum^n_{j=1} |a_{ij}|^2right)left( sum^n_{j=1} |v_j|^2right) \
=& left(sum^n_{i,j=1}|a_{ij}|^2 right)|v|^2
end{align}
which means $|Av| leq C|v|$ where $C$ is independent of $v$.
answered Dec 31 '18 at 9:18
Jacky ChongJacky Chong
17.8k21128
$endgroup$
$begingroup$
The norm given is an arbitrary norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 9:41
$begingroup$
This answer is wrong. It is valid for the usual norm on Euclidean spaces, not for a general norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:10
$begingroup$
@KaviRamaMurthy I think the answer is okay. Observe $|v_1-w_1| = |v_1 mathbf{e}_1 - w_1mathbf{e}_1| = |P_1 v-P_1w|leq |P_1|_text{op} |v-w|$
$endgroup$
– Jacky Chong
Dec 31 '18 at 23:46
$begingroup$
You have not understood the question. you are assuming what you are asked to prove. You are asked to show that $P_1$ is a continuous map. How can you assume that it is a bounded operator to prove that it is continuous? @Jacky Chong
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 23:58
$begingroup$
@KaviRamaMurthy I guess I'm assuming all linear maps on finite dimensional vector spaces are bounded.
$endgroup$
– Jacky Chong
Jan 1 at 0:03
|
show 5 more comments
$begingroup$
Hint: Let $P_1 v = v_1$. Observe
begin{align}
|v_1-w_1| = |P_1v-P_1w| leq |P_1||v-w|.
end{align}
Edit: A upon the request of @KaviRamaMurthy, I have decided to include a quick proof that all linear maps $A:mathbb{R}^n rightarrow mathbb{R}^n$ are bounded.
Using matrix representation, we see that
begin{align}
|Av|^2=& sum^n_{j=1}left|sum^n_{j=1} a_{ij}v_j right|^2 leq sum^n_{j=1}left( sum^n_{j=1} |a_{ij}|^2right)left( sum^n_{j=1} |v_j|^2right) \
=& left(sum^n_{i,j=1}|a_{ij}|^2 right)|v|^2
end{align}
which means $|Av| leq C|v|$ where $C$ is independent of $v$.
answered Dec 31 '18 at 9:18
Jacky ChongJacky Chong
17.8k21128
$endgroup$
Hint: Let $P_1 v = v_1$. Observe
begin{align}
|v_1-w_1| = |P_1v-P_1w| leq |P_1||v-w|.
end{align}
Edit: A upon the request of @KaviRamaMurthy, I have decided to include a quick proof that all linear maps $A:mathbb{R}^n rightarrow mathbb{R}^n$ are bounded.
Using matrix representation, we see that
begin{align}
|Av|^2=& sum^n_{j=1}left|sum^n_{j=1} a_{ij}v_j right|^2 leq sum^n_{j=1}left( sum^n_{j=1} |a_{ij}|^2right)left( sum^n_{j=1} |v_j|^2right) \
=& left(sum^n_{i,j=1}|a_{ij}|^2 right)|v|^2
end{align}
which means $|Av| leq C|v|$ where $C$ is independent of $v$.
answered Dec 31 '18 at 9:18
Jacky ChongJacky Chong
17.8k21128
edited Jan 1 at 0:18
answered Dec 31 '18 at 9:18
Jacky ChongJacky Chong
17.8k21128
answered Dec 31 '18 at 9:18
Jacky ChongJacky Chong
17.8k21128
answered Dec 31 '18 at 9:18
Jacky ChongJacky Chong
17.8k21128
17.8k21128
$begingroup$
The norm given is an arbitrary norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 9:41
$begingroup$
This answer is wrong. It is valid for the usual norm on Euclidean spaces, not for a general norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:10
$begingroup$
@KaviRamaMurthy I think the answer is okay. Observe $|v_1-w_1| = |v_1 mathbf{e}_1 - w_1mathbf{e}_1| = |P_1 v-P_1w|leq |P_1|_text{op} |v-w|$
$endgroup$
– Jacky Chong
Dec 31 '18 at 23:46
$begingroup$
You have not understood the question. you are assuming what you are asked to prove. You are asked to show that $P_1$ is a continuous map. How can you assume that it is a bounded operator to prove that it is continuous? @Jacky Chong
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 23:58
$begingroup$
@KaviRamaMurthy I guess I'm assuming all linear maps on finite dimensional vector spaces are bounded.
$endgroup$
– Jacky Chong
Jan 1 at 0:03
|
show 5 more comments
$begingroup$
The norm given is an arbitrary norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 9:41
$begingroup$
This answer is wrong. It is valid for the usual norm on Euclidean spaces, not for a general norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:10
$begingroup$
@KaviRamaMurthy I think the answer is okay. Observe $|v_1-w_1| = |v_1 mathbf{e}_1 - w_1mathbf{e}_1| = |P_1 v-P_1w|leq |P_1|_text{op} |v-w|$
$endgroup$
– Jacky Chong
Dec 31 '18 at 23:46
$begingroup$
You have not understood the question. you are assuming what you are asked to prove. You are asked to show that $P_1$ is a continuous map. How can you assume that it is a bounded operator to prove that it is continuous? @Jacky Chong
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 23:58
$begingroup$
@KaviRamaMurthy I guess I'm assuming all linear maps on finite dimensional vector spaces are bounded.
$endgroup$
– Jacky Chong
Jan 1 at 0:03
$begingroup$
The norm given is an arbitrary norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 9:41
$begingroup$
The norm given is an arbitrary norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 9:41
$begingroup$
This answer is wrong. It is valid for the usual norm on Euclidean spaces, not for a general norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:10
$begingroup$
This answer is wrong. It is valid for the usual norm on Euclidean spaces, not for a general norm.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:10
$begingroup$
@KaviRamaMurthy I think the answer is okay. Observe $|v_1-w_1| = |v_1 mathbf{e}_1 - w_1mathbf{e}_1| = |P_1 v-P_1w|leq |P_1|_text{op} |v-w|$
$endgroup$
– Jacky Chong
Dec 31 '18 at 23:46
$begingroup$
@KaviRamaMurthy I think the answer is okay. Observe $|v_1-w_1| = |v_1 mathbf{e}_1 - w_1mathbf{e}_1| = |P_1 v-P_1w|leq |P_1|_text{op} |v-w|$
$endgroup$
– Jacky Chong
Dec 31 '18 at 23:46
$begingroup$
You have not understood the question. you are assuming what you are asked to prove. You are asked to show that $P_1$ is a continuous map. How can you assume that it is a bounded operator to prove that it is continuous? @Jacky Chong
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 23:58
$begingroup$
You have not understood the question. you are assuming what you are asked to prove. You are asked to show that $P_1$ is a continuous map. How can you assume that it is a bounded operator to prove that it is continuous? @Jacky Chong
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 23:58
$begingroup$
@KaviRamaMurthy I guess I'm assuming all linear maps on finite dimensional vector spaces are bounded.
$endgroup$
– Jacky Chong
Jan 1 at 0:03
$begingroup$
@KaviRamaMurthy I guess I'm assuming all linear maps on finite dimensional vector spaces are bounded.
$endgroup$
– Jacky Chong
Jan 1 at 0:03
|
show 5 more comments
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$begingroup$
The norm in $X$ induces a topology, namely open sets.
$endgroup$
– dmtri
Dec 31 '18 at 9:36
$begingroup$
There are several theorems about finite dimensional normed linear spaces: any two norms are equivalent. any linear map on then is continuous etc. [These two theorems are easily seen to be equivalent]. They provide an answer to your question]. But if you want to avoid these theorems and use only the definition of norm you can look at my answer. @Jiu
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:08
$begingroup$
@KaviRamaMurthy In fact I was trying to prove that any two norms on a finite dimensional space are equivalent. And I needed the continuity of $f$ for my proof.
$endgroup$
– Jiu
Dec 31 '18 at 10:11