Show that there is some linear function on any infinite dimensional banach space that is unbounded.
$begingroup$
Let $(V,||cdot||)$ be an infinite dimensional Banach Space. Show that there is some linear function $phi:V rightarrow mathbb{R}$ that is unbounded.
I have seen a proof somewhere. It goes like this: Let $B$ be an uncountable basis for $V$, take any countable subset ${e_n: n in mathbb{N}}$ from $B$ and define $f(e_n)=n||e_n||$ and $f(x)=0$ for other basis vectors.
I have some questions about this proof.
(1) Can we always extract a countable subset from a uncountable set?
(2) I don't think $f$ is linear. $f(e_i + e_j)=n||e_i+e_j||$ which may not be equal to $n||e_i||+n||e_j||$.
real-analysis linear-algebra functional-analysis
asked Nov 30 '18 at 12:57
bbwbbw
47038
$endgroup$
add a comment |
$begingroup$
Let $(V,||cdot||)$ be an infinite dimensional Banach Space. Show that there is some linear function $phi:V rightarrow mathbb{R}$ that is unbounded.
I have seen a proof somewhere. It goes like this: Let $B$ be an uncountable basis for $V$, take any countable subset ${e_n: n in mathbb{N}}$ from $B$ and define $f(e_n)=n||e_n||$ and $f(x)=0$ for other basis vectors.
I have some questions about this proof.
(1) Can we always extract a countable subset from a uncountable set?
(2) I don't think $f$ is linear. $f(e_i + e_j)=n||e_i+e_j||$ which may not be equal to $n||e_i||+n||e_j||$.
real-analysis linear-algebra functional-analysis
asked Nov 30 '18 at 12:57
bbwbbw
47038
$endgroup$
$begingroup$
the definition isn't $f(x) = n|x|$ which wouldn't be linear, its $f(e_n) = n|e_n|$ and then linearly extended so that $f(x e_n) = x n |e_n|$ and you do this for every $e_n$
$endgroup$
– Calvin Khor
Nov 30 '18 at 13:16
$begingroup$
@CalvinKhor, sorry I'm not sure what you mean. I did wrote $f(e_n)=n||e_n||$ and you said I should define $f(e_n)=n||e_n||$. What does that mean?
$endgroup$
– bbw
Nov 30 '18 at 21:13
$begingroup$
Lets try a 2D example. Say I define $e_1 = (1,0), e_2 = (1,1)$. What do you think $f(e_1 + e_2) = ?$ there is no $n$ there. $f(e_1) = 1|e_1| = 1, f(e_2) = 2 |e_2| = 2 sqrt 2$. In fact if we force $f$ to be linear, we must have $f(e_1 + e_2) = f(e_1) + f(e_2) =1|e_1| + 2 |e_2|$. What about $f(-e_1)$? it must be $-f(e_1)$ by linearity again.
$endgroup$
– Calvin Khor
Nov 30 '18 at 21:30
add a comment |
$begingroup$
Let $(V,||cdot||)$ be an infinite dimensional Banach Space. Show that there is some linear function $phi:V rightarrow mathbb{R}$ that is unbounded.
I have seen a proof somewhere. It goes like this: Let $B$ be an uncountable basis for $V$, take any countable subset ${e_n: n in mathbb{N}}$ from $B$ and define $f(e_n)=n||e_n||$ and $f(x)=0$ for other basis vectors.
I have some questions about this proof.
(1) Can we always extract a countable subset from a uncountable set?
(2) I don't think $f$ is linear. $f(e_i + e_j)=n||e_i+e_j||$ which may not be equal to $n||e_i||+n||e_j||$.
real-analysis linear-algebra functional-analysis
asked Nov 30 '18 at 12:57
bbwbbw
47038
$endgroup$
Let $(V,||cdot||)$ be an infinite dimensional Banach Space. Show that there is some linear function $phi:V rightarrow mathbb{R}$ that is unbounded.
I have seen a proof somewhere. It goes like this: Let $B$ be an uncountable basis for $V$, take any countable subset ${e_n: n in mathbb{N}}$ from $B$ and define $f(e_n)=n||e_n||$ and $f(x)=0$ for other basis vectors.
I have some questions about this proof.
(1) Can we always extract a countable subset from a uncountable set?
(2) I don't think $f$ is linear. $f(e_i + e_j)=n||e_i+e_j||$ which may not be equal to $n||e_i||+n||e_j||$.
real-analysis linear-algebra functional-analysis
real-analysis linear-algebra functional-analysis
asked Nov 30 '18 at 12:57
bbwbbw
47038
asked Nov 30 '18 at 12:57
bbwbbw
47038
asked Nov 30 '18 at 12:57
bbwbbw
47038
asked Nov 30 '18 at 12:57
bbwbbw
47038
asked Nov 30 '18 at 12:57
bbwbbw
47038
47038
$begingroup$
the definition isn't $f(x) = n|x|$ which wouldn't be linear, its $f(e_n) = n|e_n|$ and then linearly extended so that $f(x e_n) = x n |e_n|$ and you do this for every $e_n$
$endgroup$
– Calvin Khor
Nov 30 '18 at 13:16
$begingroup$
@CalvinKhor, sorry I'm not sure what you mean. I did wrote $f(e_n)=n||e_n||$ and you said I should define $f(e_n)=n||e_n||$. What does that mean?
$endgroup$
– bbw
Nov 30 '18 at 21:13
$begingroup$
Lets try a 2D example. Say I define $e_1 = (1,0), e_2 = (1,1)$. What do you think $f(e_1 + e_2) = ?$ there is no $n$ there. $f(e_1) = 1|e_1| = 1, f(e_2) = 2 |e_2| = 2 sqrt 2$. In fact if we force $f$ to be linear, we must have $f(e_1 + e_2) = f(e_1) + f(e_2) =1|e_1| + 2 |e_2|$. What about $f(-e_1)$? it must be $-f(e_1)$ by linearity again.
$endgroup$
– Calvin Khor
Nov 30 '18 at 21:30
add a comment |
$begingroup$
the definition isn't $f(x) = n|x|$ which wouldn't be linear, its $f(e_n) = n|e_n|$ and then linearly extended so that $f(x e_n) = x n |e_n|$ and you do this for every $e_n$
$endgroup$
– Calvin Khor
Nov 30 '18 at 13:16
$begingroup$
@CalvinKhor, sorry I'm not sure what you mean. I did wrote $f(e_n)=n||e_n||$ and you said I should define $f(e_n)=n||e_n||$. What does that mean?
$endgroup$
– bbw
Nov 30 '18 at 21:13
$begingroup$
Lets try a 2D example. Say I define $e_1 = (1,0), e_2 = (1,1)$. What do you think $f(e_1 + e_2) = ?$ there is no $n$ there. $f(e_1) = 1|e_1| = 1, f(e_2) = 2 |e_2| = 2 sqrt 2$. In fact if we force $f$ to be linear, we must have $f(e_1 + e_2) = f(e_1) + f(e_2) =1|e_1| + 2 |e_2|$. What about $f(-e_1)$? it must be $-f(e_1)$ by linearity again.
$endgroup$
– Calvin Khor
Nov 30 '18 at 21:30
$begingroup$
the definition isn't $f(x) = n|x|$ which wouldn't be linear, its $f(e_n) = n|e_n|$ and then linearly extended so that $f(x e_n) = x n |e_n|$ and you do this for every $e_n$
$endgroup$
– Calvin Khor
Nov 30 '18 at 13:16
$begingroup$
the definition isn't $f(x) = n|x|$ which wouldn't be linear, its $f(e_n) = n|e_n|$ and then linearly extended so that $f(x e_n) = x n |e_n|$ and you do this for every $e_n$
$endgroup$
– Calvin Khor
Nov 30 '18 at 13:16
$begingroup$
@CalvinKhor, sorry I'm not sure what you mean. I did wrote $f(e_n)=n||e_n||$ and you said I should define $f(e_n)=n||e_n||$. What does that mean?
$endgroup$
– bbw
Nov 30 '18 at 21:13
$begingroup$
@CalvinKhor, sorry I'm not sure what you mean. I did wrote $f(e_n)=n||e_n||$ and you said I should define $f(e_n)=n||e_n||$. What does that mean?
$endgroup$
– bbw
Nov 30 '18 at 21:13
$begingroup$
Lets try a 2D example. Say I define $e_1 = (1,0), e_2 = (1,1)$. What do you think $f(e_1 + e_2) = ?$ there is no $n$ there. $f(e_1) = 1|e_1| = 1, f(e_2) = 2 |e_2| = 2 sqrt 2$. In fact if we force $f$ to be linear, we must have $f(e_1 + e_2) = f(e_1) + f(e_2) =1|e_1| + 2 |e_2|$. What about $f(-e_1)$? it must be $-f(e_1)$ by linearity again.
$endgroup$
– Calvin Khor
Nov 30 '18 at 21:30
$begingroup$
Lets try a 2D example. Say I define $e_1 = (1,0), e_2 = (1,1)$. What do you think $f(e_1 + e_2) = ?$ there is no $n$ there. $f(e_1) = 1|e_1| = 1, f(e_2) = 2 |e_2| = 2 sqrt 2$. In fact if we force $f$ to be linear, we must have $f(e_1 + e_2) = f(e_1) + f(e_2) =1|e_1| + 2 |e_2|$. What about $f(-e_1)$? it must be $-f(e_1)$ by linearity again.
$endgroup$
– Calvin Khor
Nov 30 '18 at 21:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
1) well, as soon as you accept a basis on your infinite dimensional vectorspace, you have to accept the axiom of choice, which gives you such a subset.
2) it is defined on the basis, this means that you extend by linearity, i.e. $$v=sum_{i=1}^n alpha_i v_i mapsto sum_{i=1}^n alpha_i f(v_i)$$
and so this is clearly linear. It is something that is used quite often and should be usual for you if you are already at banachspaces. If not, I recommend you revise that.
answered Nov 30 '18 at 13:11
EnkiduEnkidu
1,28319
$endgroup$
$begingroup$
But I did define it on the basis, didn't I? I defined $f(e_n)=n||e_n||$ and $f=0$ on other basis vectors right? And what do you mean by "extend by linearity"?
$endgroup$
– bbw
Nov 30 '18 at 21:17
add a comment |
$begingroup$
(1) It can be constructed as the limit of finite subsets
(2) It defines the evaluation only of the basis, I guess "extend by linearity" is implied.
answered Nov 30 '18 at 13:06
Eduardo ElaelEduardo Elael
30115
$endgroup$
add a comment |
$begingroup$
$B$ has one element, call it $b_1$. $B_1 = Bsetminus{{b_1}}$ is not empty, since otherwise $B$ is finite. $B_1$ has one element, call it $b_2$. $B_2 = B_1setminus{{b_2}}$ is similarly not empty. Inductively we construct an infinite sequence $b_1,b_2,dots$ in $B$.- Let $f_1,f_2$ be any numbers, Let $e_1,e_2 $ be linearly independent in a dimension 2 vector space $V$, and $f:{e_1,e_2}to {f_1,f_2 } $ be defined by
$$ f(e_i) = f_i.$$ You should verify that if $x$ has the basis expansion $x=x_1 e_1 + x_2e_2$,
$$ F: Vto mathbb R,quad F(x) := x_1 f_1 + x_2 f_2$$
is a well-defined linear function. Moreover, you should check that $F=f$ on the common domain of definition of two points ${e_1,e_2}$. $F$ is called the linear extension of $f$. - Exercise: extend the above definition to any dimension...
Remark - note that when you said
$$ f(e_i + e_j) = n |e_i + e_j|$$
$n$ makes no sense at all. The $n$ is
the index of some basis vector. You seem to be treating the definition as if you know what $n$ is, and then define $f(x) = n|x|$. This function is clearly not linear, yes, but this is not related to the linear extension of the function $$f:{e_1,e_2,dots}tomathbb R, quad f(e_🐶) := 🐶|e_🐶|.$$
Second remark - when you define $f$ to be zero for 'other basis vectors', this refers to the basis vectors in $Bsetminus {e_1,e_2,dots}$.
answered Nov 30 '18 at 21:50
Calvin KhorCalvin Khor
11.3k21438
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
1) well, as soon as you accept a basis on your infinite dimensional vectorspace, you have to accept the axiom of choice, which gives you such a subset.
2) it is defined on the basis, this means that you extend by linearity, i.e. $$v=sum_{i=1}^n alpha_i v_i mapsto sum_{i=1}^n alpha_i f(v_i)$$
and so this is clearly linear. It is something that is used quite often and should be usual for you if you are already at banachspaces. If not, I recommend you revise that.
answered Nov 30 '18 at 13:11
EnkiduEnkidu
1,28319
$endgroup$
$begingroup$
But I did define it on the basis, didn't I? I defined $f(e_n)=n||e_n||$ and $f=0$ on other basis vectors right? And what do you mean by "extend by linearity"?
$endgroup$
– bbw
Nov 30 '18 at 21:17
add a comment |
$begingroup$
1) well, as soon as you accept a basis on your infinite dimensional vectorspace, you have to accept the axiom of choice, which gives you such a subset.
2) it is defined on the basis, this means that you extend by linearity, i.e. $$v=sum_{i=1}^n alpha_i v_i mapsto sum_{i=1}^n alpha_i f(v_i)$$
and so this is clearly linear. It is something that is used quite often and should be usual for you if you are already at banachspaces. If not, I recommend you revise that.
answered Nov 30 '18 at 13:11
EnkiduEnkidu
1,28319
$endgroup$
$begingroup$
But I did define it on the basis, didn't I? I defined $f(e_n)=n||e_n||$ and $f=0$ on other basis vectors right? And what do you mean by "extend by linearity"?
$endgroup$
– bbw
Nov 30 '18 at 21:17
add a comment |
$begingroup$
1) well, as soon as you accept a basis on your infinite dimensional vectorspace, you have to accept the axiom of choice, which gives you such a subset.
2) it is defined on the basis, this means that you extend by linearity, i.e. $$v=sum_{i=1}^n alpha_i v_i mapsto sum_{i=1}^n alpha_i f(v_i)$$
and so this is clearly linear. It is something that is used quite often and should be usual for you if you are already at banachspaces. If not, I recommend you revise that.
answered Nov 30 '18 at 13:11
EnkiduEnkidu
1,28319
$endgroup$
1) well, as soon as you accept a basis on your infinite dimensional vectorspace, you have to accept the axiom of choice, which gives you such a subset.
2) it is defined on the basis, this means that you extend by linearity, i.e. $$v=sum_{i=1}^n alpha_i v_i mapsto sum_{i=1}^n alpha_i f(v_i)$$
and so this is clearly linear. It is something that is used quite often and should be usual for you if you are already at banachspaces. If not, I recommend you revise that.
answered Nov 30 '18 at 13:11
EnkiduEnkidu
1,28319
answered Nov 30 '18 at 13:11
EnkiduEnkidu
1,28319
answered Nov 30 '18 at 13:11
EnkiduEnkidu
1,28319
answered Nov 30 '18 at 13:11
EnkiduEnkidu
1,28319
1,28319
$begingroup$
But I did define it on the basis, didn't I? I defined $f(e_n)=n||e_n||$ and $f=0$ on other basis vectors right? And what do you mean by "extend by linearity"?
$endgroup$
– bbw
Nov 30 '18 at 21:17
add a comment |
$begingroup$
But I did define it on the basis, didn't I? I defined $f(e_n)=n||e_n||$ and $f=0$ on other basis vectors right? And what do you mean by "extend by linearity"?
$endgroup$
– bbw
Nov 30 '18 at 21:17
$begingroup$
But I did define it on the basis, didn't I? I defined $f(e_n)=n||e_n||$ and $f=0$ on other basis vectors right? And what do you mean by "extend by linearity"?
$endgroup$
– bbw
Nov 30 '18 at 21:17
$begingroup$
But I did define it on the basis, didn't I? I defined $f(e_n)=n||e_n||$ and $f=0$ on other basis vectors right? And what do you mean by "extend by linearity"?
$endgroup$
– bbw
Nov 30 '18 at 21:17
add a comment |
$begingroup$
(1) It can be constructed as the limit of finite subsets
(2) It defines the evaluation only of the basis, I guess "extend by linearity" is implied.
answered Nov 30 '18 at 13:06
Eduardo ElaelEduardo Elael
30115
$endgroup$
add a comment |
$begingroup$
(1) It can be constructed as the limit of finite subsets
(2) It defines the evaluation only of the basis, I guess "extend by linearity" is implied.
answered Nov 30 '18 at 13:06
Eduardo ElaelEduardo Elael
30115
$endgroup$
add a comment |
$begingroup$
(1) It can be constructed as the limit of finite subsets
(2) It defines the evaluation only of the basis, I guess "extend by linearity" is implied.
answered Nov 30 '18 at 13:06
Eduardo ElaelEduardo Elael
30115
$endgroup$
(1) It can be constructed as the limit of finite subsets
(2) It defines the evaluation only of the basis, I guess "extend by linearity" is implied.
answered Nov 30 '18 at 13:06
Eduardo ElaelEduardo Elael
30115
answered Nov 30 '18 at 13:06
Eduardo ElaelEduardo Elael
30115
answered Nov 30 '18 at 13:06
Eduardo ElaelEduardo Elael
30115
answered Nov 30 '18 at 13:06
Eduardo ElaelEduardo Elael
30115
30115
add a comment |
add a comment |
$begingroup$
$B$ has one element, call it $b_1$. $B_1 = Bsetminus{{b_1}}$ is not empty, since otherwise $B$ is finite. $B_1$ has one element, call it $b_2$. $B_2 = B_1setminus{{b_2}}$ is similarly not empty. Inductively we construct an infinite sequence $b_1,b_2,dots$ in $B$.- Let $f_1,f_2$ be any numbers, Let $e_1,e_2 $ be linearly independent in a dimension 2 vector space $V$, and $f:{e_1,e_2}to {f_1,f_2 } $ be defined by
$$ f(e_i) = f_i.$$ You should verify that if $x$ has the basis expansion $x=x_1 e_1 + x_2e_2$,
$$ F: Vto mathbb R,quad F(x) := x_1 f_1 + x_2 f_2$$
is a well-defined linear function. Moreover, you should check that $F=f$ on the common domain of definition of two points ${e_1,e_2}$. $F$ is called the linear extension of $f$. - Exercise: extend the above definition to any dimension...
Remark - note that when you said
$$ f(e_i + e_j) = n |e_i + e_j|$$
$n$ makes no sense at all. The $n$ is
the index of some basis vector. You seem to be treating the definition as if you know what $n$ is, and then define $f(x) = n|x|$. This function is clearly not linear, yes, but this is not related to the linear extension of the function $$f:{e_1,e_2,dots}tomathbb R, quad f(e_🐶) := 🐶|e_🐶|.$$
Second remark - when you define $f$ to be zero for 'other basis vectors', this refers to the basis vectors in $Bsetminus {e_1,e_2,dots}$.
answered Nov 30 '18 at 21:50
Calvin KhorCalvin Khor
11.3k21438
$endgroup$
add a comment |
$begingroup$
$B$ has one element, call it $b_1$. $B_1 = Bsetminus{{b_1}}$ is not empty, since otherwise $B$ is finite. $B_1$ has one element, call it $b_2$. $B_2 = B_1setminus{{b_2}}$ is similarly not empty. Inductively we construct an infinite sequence $b_1,b_2,dots$ in $B$.- Let $f_1,f_2$ be any numbers, Let $e_1,e_2 $ be linearly independent in a dimension 2 vector space $V$, and $f:{e_1,e_2}to {f_1,f_2 } $ be defined by
$$ f(e_i) = f_i.$$ You should verify that if $x$ has the basis expansion $x=x_1 e_1 + x_2e_2$,
$$ F: Vto mathbb R,quad F(x) := x_1 f_1 + x_2 f_2$$
is a well-defined linear function. Moreover, you should check that $F=f$ on the common domain of definition of two points ${e_1,e_2}$. $F$ is called the linear extension of $f$. - Exercise: extend the above definition to any dimension...
Remark - note that when you said
$$ f(e_i + e_j) = n |e_i + e_j|$$
$n$ makes no sense at all. The $n$ is
the index of some basis vector. You seem to be treating the definition as if you know what $n$ is, and then define $f(x) = n|x|$. This function is clearly not linear, yes, but this is not related to the linear extension of the function $$f:{e_1,e_2,dots}tomathbb R, quad f(e_🐶) := 🐶|e_🐶|.$$
Second remark - when you define $f$ to be zero for 'other basis vectors', this refers to the basis vectors in $Bsetminus {e_1,e_2,dots}$.
answered Nov 30 '18 at 21:50
Calvin KhorCalvin Khor
11.3k21438
$endgroup$
add a comment |
$begingroup$
$B$ has one element, call it $b_1$. $B_1 = Bsetminus{{b_1}}$ is not empty, since otherwise $B$ is finite. $B_1$ has one element, call it $b_2$. $B_2 = B_1setminus{{b_2}}$ is similarly not empty. Inductively we construct an infinite sequence $b_1,b_2,dots$ in $B$.- Let $f_1,f_2$ be any numbers, Let $e_1,e_2 $ be linearly independent in a dimension 2 vector space $V$, and $f:{e_1,e_2}to {f_1,f_2 } $ be defined by
$$ f(e_i) = f_i.$$ You should verify that if $x$ has the basis expansion $x=x_1 e_1 + x_2e_2$,
$$ F: Vto mathbb R,quad F(x) := x_1 f_1 + x_2 f_2$$
is a well-defined linear function. Moreover, you should check that $F=f$ on the common domain of definition of two points ${e_1,e_2}$. $F$ is called the linear extension of $f$. - Exercise: extend the above definition to any dimension...
Remark - note that when you said
$$ f(e_i + e_j) = n |e_i + e_j|$$
$n$ makes no sense at all. The $n$ is
the index of some basis vector. You seem to be treating the definition as if you know what $n$ is, and then define $f(x) = n|x|$. This function is clearly not linear, yes, but this is not related to the linear extension of the function $$f:{e_1,e_2,dots}tomathbb R, quad f(e_🐶) := 🐶|e_🐶|.$$
Second remark - when you define $f$ to be zero for 'other basis vectors', this refers to the basis vectors in $Bsetminus {e_1,e_2,dots}$.
answered Nov 30 '18 at 21:50
Calvin KhorCalvin Khor
11.3k21438
$endgroup$
$B$ has one element, call it $b_1$. $B_1 = Bsetminus{{b_1}}$ is not empty, since otherwise $B$ is finite. $B_1$ has one element, call it $b_2$. $B_2 = B_1setminus{{b_2}}$ is similarly not empty. Inductively we construct an infinite sequence $b_1,b_2,dots$ in $B$.- Let $f_1,f_2$ be any numbers, Let $e_1,e_2 $ be linearly independent in a dimension 2 vector space $V$, and $f:{e_1,e_2}to {f_1,f_2 } $ be defined by
$$ f(e_i) = f_i.$$ You should verify that if $x$ has the basis expansion $x=x_1 e_1 + x_2e_2$,
$$ F: Vto mathbb R,quad F(x) := x_1 f_1 + x_2 f_2$$
is a well-defined linear function. Moreover, you should check that $F=f$ on the common domain of definition of two points ${e_1,e_2}$. $F$ is called the linear extension of $f$. - Exercise: extend the above definition to any dimension...
Remark - note that when you said
$$ f(e_i + e_j) = n |e_i + e_j|$$
$n$ makes no sense at all. The $n$ is
the index of some basis vector. You seem to be treating the definition as if you know what $n$ is, and then define $f(x) = n|x|$. This function is clearly not linear, yes, but this is not related to the linear extension of the function $$f:{e_1,e_2,dots}tomathbb R, quad f(e_🐶) := 🐶|e_🐶|.$$
Second remark - when you define $f$ to be zero for 'other basis vectors', this refers to the basis vectors in $Bsetminus {e_1,e_2,dots}$.
answered Nov 30 '18 at 21:50
Calvin KhorCalvin Khor
11.3k21438
edited Nov 30 '18 at 22:03
answered Nov 30 '18 at 21:50
Calvin KhorCalvin Khor
11.3k21438
answered Nov 30 '18 at 21:50
Calvin KhorCalvin Khor
11.3k21438
answered Nov 30 '18 at 21:50
Calvin KhorCalvin Khor
11.3k21438
11.3k21438
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$begingroup$
the definition isn't $f(x) = n|x|$ which wouldn't be linear, its $f(e_n) = n|e_n|$ and then linearly extended so that $f(x e_n) = x n |e_n|$ and you do this for every $e_n$
$endgroup$
– Calvin Khor
Nov 30 '18 at 13:16
$begingroup$
@CalvinKhor, sorry I'm not sure what you mean. I did wrote $f(e_n)=n||e_n||$ and you said I should define $f(e_n)=n||e_n||$. What does that mean?
$endgroup$
– bbw
Nov 30 '18 at 21:13
$begingroup$
Lets try a 2D example. Say I define $e_1 = (1,0), e_2 = (1,1)$. What do you think $f(e_1 + e_2) = ?$ there is no $n$ there. $f(e_1) = 1|e_1| = 1, f(e_2) = 2 |e_2| = 2 sqrt 2$. In fact if we force $f$ to be linear, we must have $f(e_1 + e_2) = f(e_1) + f(e_2) =1|e_1| + 2 |e_2|$. What about $f(-e_1)$? it must be $-f(e_1)$ by linearity again.
$endgroup$
– Calvin Khor
Nov 30 '18 at 21:30