Theorem 3.2. Rudin functional analysis, $f_1 leq p$ on $M_1$
I cannot understand one small implication in such theorem
Suppose
(a) $M$ is a subspace of a real vector space $X$
(b) $p : X to mathbb{R}$ satisfies $$ p(x+y) leq p(x) + p(y)
> ;text{and}; p(tx) = tp(x) $$ if $x in X$ and $y in X, t geq 0$
(c) $f : M to mathbb{R}$ is linear and $f(x) leq p(x)$ on $M$.
Then there exists a linear $Lambda : X to mathbb{R}$ such that $$
Lambda x = f(x) ; , x in M $$ and $$
-p(-x) leq Lambda x leq p(x) ;,; x in X $$
Proof:
If $M neq X$ we pick $x_1 in X$,$x notin M$ and define
$$
M_1 = left{ x + t x_1 : x in M, t in mathbb{R}right}
$$
we have $M_1$ is a vector space. Moreover we can derive
$$
f(x) + f(y) leq p(x - x_1) + p(y - x_1)
$$
We define also $alpha$ as the least upper bound of $f(x) - p(x - x_1)$ (by varying $x in M$. We can write the two inequalities
$$
begin{array}{ll}
f(x) - alpha leq p(x - x_1) & (2)\
f(x) + alpha leq p(y - x_1) & (3)
end{array}
$$
And on $M_1$ we define
$$
f_1(x + tx_1) = f(x) + talpha ;;; (4)
$$
There's this specific bit I don't get
Take $t > 0$ replace $x$ by $t^{-1}x$ in (2) and $y$ by $t^{-1}y$ in
(3), and multiply the resulting inequalities by $t$. In combination
with (4) this proves that $f_1 leq p$ on $M_1$.
How exactly doing the mentioned steps we get $f_1 leq p$ on $M_1$?
Doing the steps mentioned I got
$$
f_1(x - talpha) + f_1(y + t alpha) = f_1(x + y) leq p(x - tx_1) + p(y + t x_1),
$$
But I still struggle to arrive at the same conclusion
Thank you
functional-analysis proof-explanation
asked Nov 28 at 10:06
user8469759
1,3311616
add a comment |
I cannot understand one small implication in such theorem
Suppose
(a) $M$ is a subspace of a real vector space $X$
(b) $p : X to mathbb{R}$ satisfies $$ p(x+y) leq p(x) + p(y)
> ;text{and}; p(tx) = tp(x) $$ if $x in X$ and $y in X, t geq 0$
(c) $f : M to mathbb{R}$ is linear and $f(x) leq p(x)$ on $M$.
Then there exists a linear $Lambda : X to mathbb{R}$ such that $$
Lambda x = f(x) ; , x in M $$ and $$
-p(-x) leq Lambda x leq p(x) ;,; x in X $$
Proof:
If $M neq X$ we pick $x_1 in X$,$x notin M$ and define
$$
M_1 = left{ x + t x_1 : x in M, t in mathbb{R}right}
$$
we have $M_1$ is a vector space. Moreover we can derive
$$
f(x) + f(y) leq p(x - x_1) + p(y - x_1)
$$
We define also $alpha$ as the least upper bound of $f(x) - p(x - x_1)$ (by varying $x in M$. We can write the two inequalities
$$
begin{array}{ll}
f(x) - alpha leq p(x - x_1) & (2)\
f(x) + alpha leq p(y - x_1) & (3)
end{array}
$$
And on $M_1$ we define
$$
f_1(x + tx_1) = f(x) + talpha ;;; (4)
$$
There's this specific bit I don't get
Take $t > 0$ replace $x$ by $t^{-1}x$ in (2) and $y$ by $t^{-1}y$ in
(3), and multiply the resulting inequalities by $t$. In combination
with (4) this proves that $f_1 leq p$ on $M_1$.
How exactly doing the mentioned steps we get $f_1 leq p$ on $M_1$?
Doing the steps mentioned I got
$$
f_1(x - talpha) + f_1(y + t alpha) = f_1(x + y) leq p(x - tx_1) + p(y + t x_1),
$$
But I still struggle to arrive at the same conclusion
Thank you
functional-analysis proof-explanation
asked Nov 28 at 10:06
user8469759
1,3311616
add a comment |
I cannot understand one small implication in such theorem
Suppose
(a) $M$ is a subspace of a real vector space $X$
(b) $p : X to mathbb{R}$ satisfies $$ p(x+y) leq p(x) + p(y)
> ;text{and}; p(tx) = tp(x) $$ if $x in X$ and $y in X, t geq 0$
(c) $f : M to mathbb{R}$ is linear and $f(x) leq p(x)$ on $M$.
Then there exists a linear $Lambda : X to mathbb{R}$ such that $$
Lambda x = f(x) ; , x in M $$ and $$
-p(-x) leq Lambda x leq p(x) ;,; x in X $$
Proof:
If $M neq X$ we pick $x_1 in X$,$x notin M$ and define
$$
M_1 = left{ x + t x_1 : x in M, t in mathbb{R}right}
$$
we have $M_1$ is a vector space. Moreover we can derive
$$
f(x) + f(y) leq p(x - x_1) + p(y - x_1)
$$
We define also $alpha$ as the least upper bound of $f(x) - p(x - x_1)$ (by varying $x in M$. We can write the two inequalities
$$
begin{array}{ll}
f(x) - alpha leq p(x - x_1) & (2)\
f(x) + alpha leq p(y - x_1) & (3)
end{array}
$$
And on $M_1$ we define
$$
f_1(x + tx_1) = f(x) + talpha ;;; (4)
$$
There's this specific bit I don't get
Take $t > 0$ replace $x$ by $t^{-1}x$ in (2) and $y$ by $t^{-1}y$ in
(3), and multiply the resulting inequalities by $t$. In combination
with (4) this proves that $f_1 leq p$ on $M_1$.
How exactly doing the mentioned steps we get $f_1 leq p$ on $M_1$?
Doing the steps mentioned I got
$$
f_1(x - talpha) + f_1(y + t alpha) = f_1(x + y) leq p(x - tx_1) + p(y + t x_1),
$$
But I still struggle to arrive at the same conclusion
Thank you
functional-analysis proof-explanation
asked Nov 28 at 10:06
user8469759
1,3311616
I cannot understand one small implication in such theorem
Suppose
(a) $M$ is a subspace of a real vector space $X$
(b) $p : X to mathbb{R}$ satisfies $$ p(x+y) leq p(x) + p(y)
> ;text{and}; p(tx) = tp(x) $$ if $x in X$ and $y in X, t geq 0$
(c) $f : M to mathbb{R}$ is linear and $f(x) leq p(x)$ on $M$.
Then there exists a linear $Lambda : X to mathbb{R}$ such that $$
Lambda x = f(x) ; , x in M $$ and $$
-p(-x) leq Lambda x leq p(x) ;,; x in X $$
Proof:
If $M neq X$ we pick $x_1 in X$,$x notin M$ and define
$$
M_1 = left{ x + t x_1 : x in M, t in mathbb{R}right}
$$
we have $M_1$ is a vector space. Moreover we can derive
$$
f(x) + f(y) leq p(x - x_1) + p(y - x_1)
$$
We define also $alpha$ as the least upper bound of $f(x) - p(x - x_1)$ (by varying $x in M$. We can write the two inequalities
$$
begin{array}{ll}
f(x) - alpha leq p(x - x_1) & (2)\
f(x) + alpha leq p(y - x_1) & (3)
end{array}
$$
And on $M_1$ we define
$$
f_1(x + tx_1) = f(x) + talpha ;;; (4)
$$
There's this specific bit I don't get
Take $t > 0$ replace $x$ by $t^{-1}x$ in (2) and $y$ by $t^{-1}y$ in
(3), and multiply the resulting inequalities by $t$. In combination
with (4) this proves that $f_1 leq p$ on $M_1$.
How exactly doing the mentioned steps we get $f_1 leq p$ on $M_1$?
Doing the steps mentioned I got
$$
f_1(x - talpha) + f_1(y + t alpha) = f_1(x + y) leq p(x - tx_1) + p(y + t x_1),
$$
But I still struggle to arrive at the same conclusion
Thank you
functional-analysis proof-explanation
functional-analysis proof-explanation
asked Nov 28 at 10:06
user8469759
1,3311616
asked Nov 28 at 10:06
user8469759
1,3311616
asked Nov 28 at 10:06
user8469759
1,3311616
asked Nov 28 at 10:06
user8469759
1,3311616
asked Nov 28 at 10:06
user8469759
1,3311616
1,3311616
add a comment |
add a comment |
1 Answer
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You are unnecessarily adding the two inequalities. If you consider them separately and look at the cases $t>0$ and $t<0$ you will get $f_1 leq p$ easily.
answered Nov 28 at 10:15
Kavi Rama Murthy
48.4k31854
I missed that... Silly related question, is the purpose of this construction to prove that we can construct a space $M_1, M subset M_1$ and we can define a functional $f_1$ that extends $f$ on $M_1$? (I'm asking because the construction itself doesn't seem to be used explicitly later in the proof).
– user8469759
Nov 28 at 10:25
If you can extend it by 'one higher dimension' by this method then you can prove that maximal space to which you can get an extension is the whole of $X$. So this step is very important when you finish the proof of the theorem.
– Kavi Rama Murthy
Nov 28 at 10:28
Ok but apart from that final point (important of course), Consider when we pick the collection $mathcal{P}$ of all ordered pairs $(M',f')$ where $M'$ is a subspace containing $M$ and $f'$ extends $f$ on $M'$. That collection isn't empty because of what we just proved, right?
– user8469759
Nov 28 at 10:32
It is not empty because $(M,f)$ is already in it!
– Kavi Rama Murthy
Nov 28 at 10:34
Out of curiosity, the theorem states (in (c) ) $f(x) leq p(x)$, shouldn't it be $|f(x)| leq |p(x)|$? (I saw there's a typo in (b), I just wanna be sure (c) is correct as well).
– user8469759
Nov 29 at 10:14
add a comment |
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You are unnecessarily adding the two inequalities. If you consider them separately and look at the cases $t>0$ and $t<0$ you will get $f_1 leq p$ easily.
answered Nov 28 at 10:15
Kavi Rama Murthy
48.4k31854
I missed that... Silly related question, is the purpose of this construction to prove that we can construct a space $M_1, M subset M_1$ and we can define a functional $f_1$ that extends $f$ on $M_1$? (I'm asking because the construction itself doesn't seem to be used explicitly later in the proof).
– user8469759
Nov 28 at 10:25
If you can extend it by 'one higher dimension' by this method then you can prove that maximal space to which you can get an extension is the whole of $X$. So this step is very important when you finish the proof of the theorem.
– Kavi Rama Murthy
Nov 28 at 10:28
Ok but apart from that final point (important of course), Consider when we pick the collection $mathcal{P}$ of all ordered pairs $(M',f')$ where $M'$ is a subspace containing $M$ and $f'$ extends $f$ on $M'$. That collection isn't empty because of what we just proved, right?
– user8469759
Nov 28 at 10:32
It is not empty because $(M,f)$ is already in it!
– Kavi Rama Murthy
Nov 28 at 10:34
Out of curiosity, the theorem states (in (c) ) $f(x) leq p(x)$, shouldn't it be $|f(x)| leq |p(x)|$? (I saw there's a typo in (b), I just wanna be sure (c) is correct as well).
– user8469759
Nov 29 at 10:14
add a comment |
You are unnecessarily adding the two inequalities. If you consider them separately and look at the cases $t>0$ and $t<0$ you will get $f_1 leq p$ easily.
answered Nov 28 at 10:15
Kavi Rama Murthy
48.4k31854
I missed that... Silly related question, is the purpose of this construction to prove that we can construct a space $M_1, M subset M_1$ and we can define a functional $f_1$ that extends $f$ on $M_1$? (I'm asking because the construction itself doesn't seem to be used explicitly later in the proof).
– user8469759
Nov 28 at 10:25
If you can extend it by 'one higher dimension' by this method then you can prove that maximal space to which you can get an extension is the whole of $X$. So this step is very important when you finish the proof of the theorem.
– Kavi Rama Murthy
Nov 28 at 10:28
Ok but apart from that final point (important of course), Consider when we pick the collection $mathcal{P}$ of all ordered pairs $(M',f')$ where $M'$ is a subspace containing $M$ and $f'$ extends $f$ on $M'$. That collection isn't empty because of what we just proved, right?
– user8469759
Nov 28 at 10:32
It is not empty because $(M,f)$ is already in it!
– Kavi Rama Murthy
Nov 28 at 10:34
Out of curiosity, the theorem states (in (c) ) $f(x) leq p(x)$, shouldn't it be $|f(x)| leq |p(x)|$? (I saw there's a typo in (b), I just wanna be sure (c) is correct as well).
– user8469759
Nov 29 at 10:14
add a comment |
You are unnecessarily adding the two inequalities. If you consider them separately and look at the cases $t>0$ and $t<0$ you will get $f_1 leq p$ easily.
answered Nov 28 at 10:15
Kavi Rama Murthy
48.4k31854
You are unnecessarily adding the two inequalities. If you consider them separately and look at the cases $t>0$ and $t<0$ you will get $f_1 leq p$ easily.
answered Nov 28 at 10:15
Kavi Rama Murthy
48.4k31854
answered Nov 28 at 10:15
Kavi Rama Murthy
48.4k31854
answered Nov 28 at 10:15
Kavi Rama Murthy
48.4k31854
answered Nov 28 at 10:15
Kavi Rama Murthy
48.4k31854
48.4k31854
I missed that... Silly related question, is the purpose of this construction to prove that we can construct a space $M_1, M subset M_1$ and we can define a functional $f_1$ that extends $f$ on $M_1$? (I'm asking because the construction itself doesn't seem to be used explicitly later in the proof).
– user8469759
Nov 28 at 10:25
If you can extend it by 'one higher dimension' by this method then you can prove that maximal space to which you can get an extension is the whole of $X$. So this step is very important when you finish the proof of the theorem.
– Kavi Rama Murthy
Nov 28 at 10:28
Ok but apart from that final point (important of course), Consider when we pick the collection $mathcal{P}$ of all ordered pairs $(M',f')$ where $M'$ is a subspace containing $M$ and $f'$ extends $f$ on $M'$. That collection isn't empty because of what we just proved, right?
– user8469759
Nov 28 at 10:32
It is not empty because $(M,f)$ is already in it!
– Kavi Rama Murthy
Nov 28 at 10:34
Out of curiosity, the theorem states (in (c) ) $f(x) leq p(x)$, shouldn't it be $|f(x)| leq |p(x)|$? (I saw there's a typo in (b), I just wanna be sure (c) is correct as well).
– user8469759
Nov 29 at 10:14
add a comment |
I missed that... Silly related question, is the purpose of this construction to prove that we can construct a space $M_1, M subset M_1$ and we can define a functional $f_1$ that extends $f$ on $M_1$? (I'm asking because the construction itself doesn't seem to be used explicitly later in the proof).
– user8469759
Nov 28 at 10:25
If you can extend it by 'one higher dimension' by this method then you can prove that maximal space to which you can get an extension is the whole of $X$. So this step is very important when you finish the proof of the theorem.
– Kavi Rama Murthy
Nov 28 at 10:28
Ok but apart from that final point (important of course), Consider when we pick the collection $mathcal{P}$ of all ordered pairs $(M',f')$ where $M'$ is a subspace containing $M$ and $f'$ extends $f$ on $M'$. That collection isn't empty because of what we just proved, right?
– user8469759
Nov 28 at 10:32
It is not empty because $(M,f)$ is already in it!
– Kavi Rama Murthy
Nov 28 at 10:34
Out of curiosity, the theorem states (in (c) ) $f(x) leq p(x)$, shouldn't it be $|f(x)| leq |p(x)|$? (I saw there's a typo in (b), I just wanna be sure (c) is correct as well).
– user8469759
Nov 29 at 10:14
I missed that... Silly related question, is the purpose of this construction to prove that we can construct a space $M_1, M subset M_1$ and we can define a functional $f_1$ that extends $f$ on $M_1$? (I'm asking because the construction itself doesn't seem to be used explicitly later in the proof).
– user8469759
Nov 28 at 10:25
I missed that... Silly related question, is the purpose of this construction to prove that we can construct a space $M_1, M subset M_1$ and we can define a functional $f_1$ that extends $f$ on $M_1$? (I'm asking because the construction itself doesn't seem to be used explicitly later in the proof).
– user8469759
Nov 28 at 10:25
If you can extend it by 'one higher dimension' by this method then you can prove that maximal space to which you can get an extension is the whole of $X$. So this step is very important when you finish the proof of the theorem.
– Kavi Rama Murthy
Nov 28 at 10:28
If you can extend it by 'one higher dimension' by this method then you can prove that maximal space to which you can get an extension is the whole of $X$. So this step is very important when you finish the proof of the theorem.
– Kavi Rama Murthy
Nov 28 at 10:28
Ok but apart from that final point (important of course), Consider when we pick the collection $mathcal{P}$ of all ordered pairs $(M',f')$ where $M'$ is a subspace containing $M$ and $f'$ extends $f$ on $M'$. That collection isn't empty because of what we just proved, right?
– user8469759
Nov 28 at 10:32
Ok but apart from that final point (important of course), Consider when we pick the collection $mathcal{P}$ of all ordered pairs $(M',f')$ where $M'$ is a subspace containing $M$ and $f'$ extends $f$ on $M'$. That collection isn't empty because of what we just proved, right?
– user8469759
Nov 28 at 10:32
It is not empty because $(M,f)$ is already in it!
– Kavi Rama Murthy
Nov 28 at 10:34
It is not empty because $(M,f)$ is already in it!
– Kavi Rama Murthy
Nov 28 at 10:34
Out of curiosity, the theorem states (in (c) ) $f(x) leq p(x)$, shouldn't it be $|f(x)| leq |p(x)|$? (I saw there's a typo in (b), I just wanna be sure (c) is correct as well).
– user8469759
Nov 29 at 10:14
Out of curiosity, the theorem states (in (c) ) $f(x) leq p(x)$, shouldn't it be $|f(x)| leq |p(x)|$? (I saw there's a typo in (b), I just wanna be sure (c) is correct as well).
– user8469759
Nov 29 at 10:14
add a comment |
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