Show that there is some linear function on any infinite dimensional banach space that is unbounded.












1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














1












1








1


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










3 Answers
3






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

(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.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$



    1. $B$ 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 $B$.

    2. 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 $f$.

    3. 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 $Bsetminus {e_1,e_2,dots}$.






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        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
















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        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














      1












      1








      1





      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$



      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.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      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


















      • $begingroup$
        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











      0












      $begingroup$

      (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.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        (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.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          (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.






          share|cite|improve this answer









          $endgroup$



          (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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 13:06









          Eduardo ElaelEduardo Elael

          30115




          30115























              0












              $begingroup$



              1. $B$ 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 $B$.

              2. 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 $f$.

              3. 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 $Bsetminus {e_1,e_2,dots}$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$



                1. $B$ 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 $B$.

                2. 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 $f$.

                3. 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 $Bsetminus {e_1,e_2,dots}$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$



                  1. $B$ 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 $B$.

                  2. 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 $f$.

                  3. 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 $Bsetminus {e_1,e_2,dots}$.






                  share|cite|improve this answer











                  $endgroup$





                  1. $B$ 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 $B$.

                  2. 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 $f$.

                  3. 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 $Bsetminus {e_1,e_2,dots}$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 22:03

























                  answered Nov 30 '18 at 21:50









                  Calvin KhorCalvin Khor

                  11.3k21438




                  11.3k21438






























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