If $X$ is a finite dimensional space, then all linear functionals are bounded on $X$.












0












$begingroup$


Let $X$ be a finite dimensional space. I want to prove that all linear functionals are bounded.



MY TRIAL



Let $dim X=ngeq 1,$ and ${e_i}^{n}_{i=1}$ be a basis. Hence, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that for arbitrary $xin X,$
begin{align} x=sum^{n}_{i=1}alpha_i e_i.end{align}



Let $xin X$ be arbitrary. Then,



begin{align} |f(x)|
&=left|fleft(sum^{n}_{i=1}alpha_i e_iright)right|\&=left|sum^{n}_{i=1}alpha_i fleft(e_iright)right|\&leqsum^{n}_{i=1}left|alpha_iright| left|fleft(e_iright)right|\&leq maxlimits_{1leq ileq n}left|alpha_iright|sum^{n}_{i=1}left|fleft(e_iright)right|end{align}
Define $|cdot|_0$ as $|x|_0=maxlimits_{1leq ileq n}left|alpha_iright|$ which is a norm on $X$ [see Prove that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$.. Hence,
begin{align} |f(x)|
leq |x|_0sum^{n}_{i=1}left|fleft(e_iright)right|end{align}
But All norms defined on a finite dimensional normed linear space are equivalent, so any norm $| cdot |$ defined on $E,$ is equivalent to $| cdot |_0$, i.e.
begin{align} |f(x)|
leq |x|sum^{n}_{i=1}left|fleft(e_iright)right|=K|x|,;;text{where};K:=sum^{n}_{i=1}left|fleft(e_iright)right|geq 0,end{align}
and we are done! Please, I'm I right? Alternative proofs will be highly welcome!






functional-analysis normed-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The equivalence of norms is not quite as strong as you use here; you need an additional constant in your final inequality (but you can just absorb that constant into the definition of $K$).
    $endgroup$
    – Mees de Vries
    Dec 27 '18 at 15:57










  • $begingroup$
    @Mees de Vries: So, what do you say about the proof?
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 16:13










  • $begingroup$
    Your proof is fine, although I think it's easier if you use the $2-$ norm because then Cauchy-Schwarz gives the result almost immediately.
    $endgroup$
    – Matematleta
    Dec 27 '18 at 17:09










  • $begingroup$
    @Matematleta: That's true! Thanks for your comment!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 17:11
















0












$begingroup$


Let $X$ be a finite dimensional space. I want to prove that all linear functionals are bounded.



MY TRIAL



Let $dim X=ngeq 1,$ and ${e_i}^{n}_{i=1}$ be a basis. Hence, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that for arbitrary $xin X,$
begin{align} x=sum^{n}_{i=1}alpha_i e_i.end{align}



Let $xin X$ be arbitrary. Then,



begin{align} |f(x)|
&=left|fleft(sum^{n}_{i=1}alpha_i e_iright)right|\&=left|sum^{n}_{i=1}alpha_i fleft(e_iright)right|\&leqsum^{n}_{i=1}left|alpha_iright| left|fleft(e_iright)right|\&leq maxlimits_{1leq ileq n}left|alpha_iright|sum^{n}_{i=1}left|fleft(e_iright)right|end{align}
Define $|cdot|_0$ as $|x|_0=maxlimits_{1leq ileq n}left|alpha_iright|$ which is a norm on $X$ [see Prove that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$.. Hence,
begin{align} |f(x)|
leq |x|_0sum^{n}_{i=1}left|fleft(e_iright)right|end{align}
But All norms defined on a finite dimensional normed linear space are equivalent, so any norm $| cdot |$ defined on $E,$ is equivalent to $| cdot |_0$, i.e.
begin{align} |f(x)|
leq |x|sum^{n}_{i=1}left|fleft(e_iright)right|=K|x|,;;text{where};K:=sum^{n}_{i=1}left|fleft(e_iright)right|geq 0,end{align}
and we are done! Please, I'm I right? Alternative proofs will be highly welcome!






functional-analysis normed-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The equivalence of norms is not quite as strong as you use here; you need an additional constant in your final inequality (but you can just absorb that constant into the definition of $K$).
    $endgroup$
    – Mees de Vries
    Dec 27 '18 at 15:57










  • $begingroup$
    @Mees de Vries: So, what do you say about the proof?
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 16:13










  • $begingroup$
    Your proof is fine, although I think it's easier if you use the $2-$ norm because then Cauchy-Schwarz gives the result almost immediately.
    $endgroup$
    – Matematleta
    Dec 27 '18 at 17:09










  • $begingroup$
    @Matematleta: That's true! Thanks for your comment!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 17:11














0












0








0


1



$begingroup$


Let $X$ be a finite dimensional space. I want to prove that all linear functionals are bounded.



MY TRIAL



Let $dim X=ngeq 1,$ and ${e_i}^{n}_{i=1}$ be a basis. Hence, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that for arbitrary $xin X,$
begin{align} x=sum^{n}_{i=1}alpha_i e_i.end{align}



Let $xin X$ be arbitrary. Then,



begin{align} |f(x)|
&=left|fleft(sum^{n}_{i=1}alpha_i e_iright)right|\&=left|sum^{n}_{i=1}alpha_i fleft(e_iright)right|\&leqsum^{n}_{i=1}left|alpha_iright| left|fleft(e_iright)right|\&leq maxlimits_{1leq ileq n}left|alpha_iright|sum^{n}_{i=1}left|fleft(e_iright)right|end{align}
Define $|cdot|_0$ as $|x|_0=maxlimits_{1leq ileq n}left|alpha_iright|$ which is a norm on $X$ [see Prove that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$.. Hence,
begin{align} |f(x)|
leq |x|_0sum^{n}_{i=1}left|fleft(e_iright)right|end{align}
But All norms defined on a finite dimensional normed linear space are equivalent, so any norm $| cdot |$ defined on $E,$ is equivalent to $| cdot |_0$, i.e.
begin{align} |f(x)|
leq |x|sum^{n}_{i=1}left|fleft(e_iright)right|=K|x|,;;text{where};K:=sum^{n}_{i=1}left|fleft(e_iright)right|geq 0,end{align}
and we are done! Please, I'm I right? Alternative proofs will be highly welcome!






functional-analysis normed-spaces







share|cite|improve this question











$endgroup$




Let $X$ be a finite dimensional space. I want to prove that all linear functionals are bounded.



MY TRIAL



Let $dim X=ngeq 1,$ and ${e_i}^{n}_{i=1}$ be a basis. Hence, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that for arbitrary $xin X,$
begin{align} x=sum^{n}_{i=1}alpha_i e_i.end{align}



Let $xin X$ be arbitrary. Then,



begin{align} |f(x)|
&=left|fleft(sum^{n}_{i=1}alpha_i e_iright)right|\&=left|sum^{n}_{i=1}alpha_i fleft(e_iright)right|\&leqsum^{n}_{i=1}left|alpha_iright| left|fleft(e_iright)right|\&leq maxlimits_{1leq ileq n}left|alpha_iright|sum^{n}_{i=1}left|fleft(e_iright)right|end{align}
Define $|cdot|_0$ as $|x|_0=maxlimits_{1leq ileq n}left|alpha_iright|$ which is a norm on $X$ [see Prove that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$.. Hence,
begin{align} |f(x)|
leq |x|_0sum^{n}_{i=1}left|fleft(e_iright)right|end{align}
But All norms defined on a finite dimensional normed linear space are equivalent, so any norm $| cdot |$ defined on $E,$ is equivalent to $| cdot |_0$, i.e.
begin{align} |f(x)|
leq |x|sum^{n}_{i=1}left|fleft(e_iright)right|=K|x|,;;text{where};K:=sum^{n}_{i=1}left|fleft(e_iright)right|geq 0,end{align}
and we are done! Please, I'm I right? Alternative proofs will be highly welcome!






functional-analysis normed-spaces




functional-analysis normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 16:02







Omojola Micheal

















asked Dec 27 '18 at 15:51









Omojola MichealOmojola Micheal

2,070424




2,070424








  • 1




    $begingroup$
    The equivalence of norms is not quite as strong as you use here; you need an additional constant in your final inequality (but you can just absorb that constant into the definition of $K$).
    $endgroup$
    – Mees de Vries
    Dec 27 '18 at 15:57










  • $begingroup$
    @Mees de Vries: So, what do you say about the proof?
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 16:13










  • $begingroup$
    Your proof is fine, although I think it's easier if you use the $2-$ norm because then Cauchy-Schwarz gives the result almost immediately.
    $endgroup$
    – Matematleta
    Dec 27 '18 at 17:09










  • $begingroup$
    @Matematleta: That's true! Thanks for your comment!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 17:11














  • 1




    $begingroup$
    The equivalence of norms is not quite as strong as you use here; you need an additional constant in your final inequality (but you can just absorb that constant into the definition of $K$).
    $endgroup$
    – Mees de Vries
    Dec 27 '18 at 15:57










  • $begingroup$
    @Mees de Vries: So, what do you say about the proof?
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 16:13










  • $begingroup$
    Your proof is fine, although I think it's easier if you use the $2-$ norm because then Cauchy-Schwarz gives the result almost immediately.
    $endgroup$
    – Matematleta
    Dec 27 '18 at 17:09










  • $begingroup$
    @Matematleta: That's true! Thanks for your comment!
    $endgroup$
    – Omojola Micheal
    Dec 27 '18 at 17:11








1




1




$begingroup$
The equivalence of norms is not quite as strong as you use here; you need an additional constant in your final inequality (but you can just absorb that constant into the definition of $K$).
$endgroup$
– Mees de Vries
Dec 27 '18 at 15:57




$begingroup$
The equivalence of norms is not quite as strong as you use here; you need an additional constant in your final inequality (but you can just absorb that constant into the definition of $K$).
$endgroup$
– Mees de Vries
Dec 27 '18 at 15:57












$begingroup$
@Mees de Vries: So, what do you say about the proof?
$endgroup$
– Omojola Micheal
Dec 27 '18 at 16:13




$begingroup$
@Mees de Vries: So, what do you say about the proof?
$endgroup$
– Omojola Micheal
Dec 27 '18 at 16:13












$begingroup$
Your proof is fine, although I think it's easier if you use the $2-$ norm because then Cauchy-Schwarz gives the result almost immediately.
$endgroup$
– Matematleta
Dec 27 '18 at 17:09




$begingroup$
Your proof is fine, although I think it's easier if you use the $2-$ norm because then Cauchy-Schwarz gives the result almost immediately.
$endgroup$
– Matematleta
Dec 27 '18 at 17:09












$begingroup$
@Matematleta: That's true! Thanks for your comment!
$endgroup$
– Omojola Micheal
Dec 27 '18 at 17:11




$begingroup$
@Matematleta: That's true! Thanks for your comment!
$endgroup$
– Omojola Micheal
Dec 27 '18 at 17:11










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