Non total orthonormal set in a non Hilbert inner product space












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Suppose there exist a subset $M$ of an inner product space $X$, and the orthogonal complement of $M $ is the zero vector. If $X $ is a Hilbert Space then the span of $M $ will be dense in $X $, but this is not always true if $X $ is not a complete inner product space. Are there any examples of an orthonormal set in a non complete inner product space such that the orthorgonal complement of this set consists only of the zero vector, but this set is not a total orthonormal set? Thank you in advance.






hilbert-spaces inner-product-space orthonormal







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$endgroup$












  • $begingroup$
    Are you looking for an orthonormal set in an inner product space with orthonormal complement ${0}$ but which does not have a dense linear span? Because I doubt the existence of such a set. I'm not certain though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 18:18










  • $begingroup$
    Any orthonormal set $E $ in any inner product space $X $ thats non Hilbert, such that the orthogonal complement of $E $, denoted by $E^c $ is the zero vector and there exists no other $xin X $ that is in $E^c $ and span $E $ is not dense in $X $. Just a thought when reading Functional Analysis by Kreyszig.
    $endgroup$
    – Jonathan Ng
    Dec 12 '18 at 18:31
















1












$begingroup$


Suppose there exist a subset $M$ of an inner product space $X$, and the orthogonal complement of $M $ is the zero vector. If $X $ is a Hilbert Space then the span of $M $ will be dense in $X $, but this is not always true if $X $ is not a complete inner product space. Are there any examples of an orthonormal set in a non complete inner product space such that the orthorgonal complement of this set consists only of the zero vector, but this set is not a total orthonormal set? Thank you in advance.






hilbert-spaces inner-product-space orthonormal







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you looking for an orthonormal set in an inner product space with orthonormal complement ${0}$ but which does not have a dense linear span? Because I doubt the existence of such a set. I'm not certain though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 18:18










  • $begingroup$
    Any orthonormal set $E $ in any inner product space $X $ thats non Hilbert, such that the orthogonal complement of $E $, denoted by $E^c $ is the zero vector and there exists no other $xin X $ that is in $E^c $ and span $E $ is not dense in $X $. Just a thought when reading Functional Analysis by Kreyszig.
    $endgroup$
    – Jonathan Ng
    Dec 12 '18 at 18:31














1












1








1





$begingroup$


Suppose there exist a subset $M$ of an inner product space $X$, and the orthogonal complement of $M $ is the zero vector. If $X $ is a Hilbert Space then the span of $M $ will be dense in $X $, but this is not always true if $X $ is not a complete inner product space. Are there any examples of an orthonormal set in a non complete inner product space such that the orthorgonal complement of this set consists only of the zero vector, but this set is not a total orthonormal set? Thank you in advance.






hilbert-spaces inner-product-space orthonormal







share|cite|improve this question









$endgroup$




Suppose there exist a subset $M$ of an inner product space $X$, and the orthogonal complement of $M $ is the zero vector. If $X $ is a Hilbert Space then the span of $M $ will be dense in $X $, but this is not always true if $X $ is not a complete inner product space. Are there any examples of an orthonormal set in a non complete inner product space such that the orthorgonal complement of this set consists only of the zero vector, but this set is not a total orthonormal set? Thank you in advance.






hilbert-spaces inner-product-space orthonormal




hilbert-spaces inner-product-space orthonormal






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share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 18:08









Jonathan NgJonathan Ng

112




112












  • $begingroup$
    Are you looking for an orthonormal set in an inner product space with orthonormal complement ${0}$ but which does not have a dense linear span? Because I doubt the existence of such a set. I'm not certain though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 18:18










  • $begingroup$
    Any orthonormal set $E $ in any inner product space $X $ thats non Hilbert, such that the orthogonal complement of $E $, denoted by $E^c $ is the zero vector and there exists no other $xin X $ that is in $E^c $ and span $E $ is not dense in $X $. Just a thought when reading Functional Analysis by Kreyszig.
    $endgroup$
    – Jonathan Ng
    Dec 12 '18 at 18:31


















  • $begingroup$
    Are you looking for an orthonormal set in an inner product space with orthonormal complement ${0}$ but which does not have a dense linear span? Because I doubt the existence of such a set. I'm not certain though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 18:18










  • $begingroup$
    Any orthonormal set $E $ in any inner product space $X $ thats non Hilbert, such that the orthogonal complement of $E $, denoted by $E^c $ is the zero vector and there exists no other $xin X $ that is in $E^c $ and span $E $ is not dense in $X $. Just a thought when reading Functional Analysis by Kreyszig.
    $endgroup$
    – Jonathan Ng
    Dec 12 '18 at 18:31
















$begingroup$
Are you looking for an orthonormal set in an inner product space with orthonormal complement ${0}$ but which does not have a dense linear span? Because I doubt the existence of such a set. I'm not certain though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 18:18




$begingroup$
Are you looking for an orthonormal set in an inner product space with orthonormal complement ${0}$ but which does not have a dense linear span? Because I doubt the existence of such a set. I'm not certain though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 18:18












$begingroup$
Any orthonormal set $E $ in any inner product space $X $ thats non Hilbert, such that the orthogonal complement of $E $, denoted by $E^c $ is the zero vector and there exists no other $xin X $ that is in $E^c $ and span $E $ is not dense in $X $. Just a thought when reading Functional Analysis by Kreyszig.
$endgroup$
– Jonathan Ng
Dec 12 '18 at 18:31




$begingroup$
Any orthonormal set $E $ in any inner product space $X $ thats non Hilbert, such that the orthogonal complement of $E $, denoted by $E^c $ is the zero vector and there exists no other $xin X $ that is in $E^c $ and span $E $ is not dense in $X $. Just a thought when reading Functional Analysis by Kreyszig.
$endgroup$
– Jonathan Ng
Dec 12 '18 at 18:31










1 Answer
1






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oldest

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2












$begingroup$

Let $E$ be the space of sequences of real numbers with finite support endowed with the $l^2$ inner product and define the linear form $phi colon E to mathbb{R}$ by
$$phi(x) = sum_{n = 0}^infty frac{x_n}{n+1}, quad x in E.$$
Then $phi$ is a continuous linear form, so that $F = ker phi$ is a closed subspace. Notice that $F neq E$ (for instance take $x_0 = 1, x_1=0, x_2 = 0, ldots$). In particular $F$ is not dense in $E$. Finally we can show that $F^perp = {0}$ (see below for a more elegant solution). Indeed let $y in F^perp$ and fix $n geq 0$. Define $x in E$ by
$$begin{align*}x_n &= n+1, \
x_{n+1} &= -(n+2),\
x_k &= 0, quad k notin {n,n+1}.end{align*}$$
Then $x in F$ so that by definition of $y$ we have $0 = langle x,y rangle = (n+1)y_n - (n+2)y_{n+1}$, i.e. $$y_{n+1} = frac{n+1}{n+2} , y_n.$$
This is true for all $ngeq 0$. Since we assumed that $y$ has finite support, this is only possible if $y equiv 0$.




As suggested in the comment by Hanno, another way to prove that $F^perp = {0}$ is the following: notice that $phi(x) = langle x , a rangle$ with $a = left( frac{1}{n+1} right)_{n geq 0} in l^2- E$. Then in $l^2$, we have $ker phi = {a}^perp$ and $(ker phi)^perp = text{span}{a}$, so $F^perp = text{span}{a} cap E = {0}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Let me state two points (1) A bit misleading: You write "Notice that $Fneq E$", but the $x$ given inside the parentheses illustrates $kerphine{0}$. (2) Conceptual: The linear form is $phi(x) = langle x,a rangle$ with fixed $a = left(frac 1{n+1}right)_{n=0,1,dots}in ell^2(mathbb N_0)backslash E,$. Then in $ell^2(mathbb N_0)$ one has $(kerphi)^perp = langle arangle$, and $langle aranglecap E ={0}$.
    $endgroup$
    – Hanno
    Dec 29 '18 at 9:55












  • $begingroup$
    @Hanno (1) Yes you are correct about that and I will edit my post. (2) I chose not to talk about $l^2$ because in my eyes it is more confusing than anything else, but again you are correct and it is definitely a more elegant solution. Thanks for the remarks.
    $endgroup$
    – Michh
    Dec 29 '18 at 11:04










  • $begingroup$
    The title of the question would suggest that the OP is looking for an orthonormal subset with this property, not an entire subspace. Can you find an orthonormal basis for $F$? It exists by Gram-Schmidt but that isn't really explicit.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 11:42








  • 1




    $begingroup$
    @mechanodroid Reading off from the answer, the set ${e_n-frac{n+2}{n+1}e_{n+1}mid ninmathbb N_0}$ is a basis for F. Running Gram-Schmidt would yield an orthonormal one, but I do not feel any necessity to carry this through.
    $endgroup$
    – Hanno
    Dec 29 '18 at 12:03










  • $begingroup$
    @Hanno You're right, I agree.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 12:29











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1 Answer
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2












$begingroup$

Let $E$ be the space of sequences of real numbers with finite support endowed with the $l^2$ inner product and define the linear form $phi colon E to mathbb{R}$ by
$$phi(x) = sum_{n = 0}^infty frac{x_n}{n+1}, quad x in E.$$
Then $phi$ is a continuous linear form, so that $F = ker phi$ is a closed subspace. Notice that $F neq E$ (for instance take $x_0 = 1, x_1=0, x_2 = 0, ldots$). In particular $F$ is not dense in $E$. Finally we can show that $F^perp = {0}$ (see below for a more elegant solution). Indeed let $y in F^perp$ and fix $n geq 0$. Define $x in E$ by
$$begin{align*}x_n &= n+1, \
x_{n+1} &= -(n+2),\
x_k &= 0, quad k notin {n,n+1}.end{align*}$$
Then $x in F$ so that by definition of $y$ we have $0 = langle x,y rangle = (n+1)y_n - (n+2)y_{n+1}$, i.e. $$y_{n+1} = frac{n+1}{n+2} , y_n.$$
This is true for all $ngeq 0$. Since we assumed that $y$ has finite support, this is only possible if $y equiv 0$.




As suggested in the comment by Hanno, another way to prove that $F^perp = {0}$ is the following: notice that $phi(x) = langle x , a rangle$ with $a = left( frac{1}{n+1} right)_{n geq 0} in l^2- E$. Then in $l^2$, we have $ker phi = {a}^perp$ and $(ker phi)^perp = text{span}{a}$, so $F^perp = text{span}{a} cap E = {0}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Let me state two points (1) A bit misleading: You write "Notice that $Fneq E$", but the $x$ given inside the parentheses illustrates $kerphine{0}$. (2) Conceptual: The linear form is $phi(x) = langle x,a rangle$ with fixed $a = left(frac 1{n+1}right)_{n=0,1,dots}in ell^2(mathbb N_0)backslash E,$. Then in $ell^2(mathbb N_0)$ one has $(kerphi)^perp = langle arangle$, and $langle aranglecap E ={0}$.
    $endgroup$
    – Hanno
    Dec 29 '18 at 9:55












  • $begingroup$
    @Hanno (1) Yes you are correct about that and I will edit my post. (2) I chose not to talk about $l^2$ because in my eyes it is more confusing than anything else, but again you are correct and it is definitely a more elegant solution. Thanks for the remarks.
    $endgroup$
    – Michh
    Dec 29 '18 at 11:04










  • $begingroup$
    The title of the question would suggest that the OP is looking for an orthonormal subset with this property, not an entire subspace. Can you find an orthonormal basis for $F$? It exists by Gram-Schmidt but that isn't really explicit.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 11:42








  • 1




    $begingroup$
    @mechanodroid Reading off from the answer, the set ${e_n-frac{n+2}{n+1}e_{n+1}mid ninmathbb N_0}$ is a basis for F. Running Gram-Schmidt would yield an orthonormal one, but I do not feel any necessity to carry this through.
    $endgroup$
    – Hanno
    Dec 29 '18 at 12:03










  • $begingroup$
    @Hanno You're right, I agree.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 12:29
















2












$begingroup$

Let $E$ be the space of sequences of real numbers with finite support endowed with the $l^2$ inner product and define the linear form $phi colon E to mathbb{R}$ by
$$phi(x) = sum_{n = 0}^infty frac{x_n}{n+1}, quad x in E.$$
Then $phi$ is a continuous linear form, so that $F = ker phi$ is a closed subspace. Notice that $F neq E$ (for instance take $x_0 = 1, x_1=0, x_2 = 0, ldots$). In particular $F$ is not dense in $E$. Finally we can show that $F^perp = {0}$ (see below for a more elegant solution). Indeed let $y in F^perp$ and fix $n geq 0$. Define $x in E$ by
$$begin{align*}x_n &= n+1, \
x_{n+1} &= -(n+2),\
x_k &= 0, quad k notin {n,n+1}.end{align*}$$
Then $x in F$ so that by definition of $y$ we have $0 = langle x,y rangle = (n+1)y_n - (n+2)y_{n+1}$, i.e. $$y_{n+1} = frac{n+1}{n+2} , y_n.$$
This is true for all $ngeq 0$. Since we assumed that $y$ has finite support, this is only possible if $y equiv 0$.




As suggested in the comment by Hanno, another way to prove that $F^perp = {0}$ is the following: notice that $phi(x) = langle x , a rangle$ with $a = left( frac{1}{n+1} right)_{n geq 0} in l^2- E$. Then in $l^2$, we have $ker phi = {a}^perp$ and $(ker phi)^perp = text{span}{a}$, so $F^perp = text{span}{a} cap E = {0}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Let me state two points (1) A bit misleading: You write "Notice that $Fneq E$", but the $x$ given inside the parentheses illustrates $kerphine{0}$. (2) Conceptual: The linear form is $phi(x) = langle x,a rangle$ with fixed $a = left(frac 1{n+1}right)_{n=0,1,dots}in ell^2(mathbb N_0)backslash E,$. Then in $ell^2(mathbb N_0)$ one has $(kerphi)^perp = langle arangle$, and $langle aranglecap E ={0}$.
    $endgroup$
    – Hanno
    Dec 29 '18 at 9:55












  • $begingroup$
    @Hanno (1) Yes you are correct about that and I will edit my post. (2) I chose not to talk about $l^2$ because in my eyes it is more confusing than anything else, but again you are correct and it is definitely a more elegant solution. Thanks for the remarks.
    $endgroup$
    – Michh
    Dec 29 '18 at 11:04










  • $begingroup$
    The title of the question would suggest that the OP is looking for an orthonormal subset with this property, not an entire subspace. Can you find an orthonormal basis for $F$? It exists by Gram-Schmidt but that isn't really explicit.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 11:42








  • 1




    $begingroup$
    @mechanodroid Reading off from the answer, the set ${e_n-frac{n+2}{n+1}e_{n+1}mid ninmathbb N_0}$ is a basis for F. Running Gram-Schmidt would yield an orthonormal one, but I do not feel any necessity to carry this through.
    $endgroup$
    – Hanno
    Dec 29 '18 at 12:03










  • $begingroup$
    @Hanno You're right, I agree.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 12:29














2












2








2





$begingroup$

Let $E$ be the space of sequences of real numbers with finite support endowed with the $l^2$ inner product and define the linear form $phi colon E to mathbb{R}$ by
$$phi(x) = sum_{n = 0}^infty frac{x_n}{n+1}, quad x in E.$$
Then $phi$ is a continuous linear form, so that $F = ker phi$ is a closed subspace. Notice that $F neq E$ (for instance take $x_0 = 1, x_1=0, x_2 = 0, ldots$). In particular $F$ is not dense in $E$. Finally we can show that $F^perp = {0}$ (see below for a more elegant solution). Indeed let $y in F^perp$ and fix $n geq 0$. Define $x in E$ by
$$begin{align*}x_n &= n+1, \
x_{n+1} &= -(n+2),\
x_k &= 0, quad k notin {n,n+1}.end{align*}$$
Then $x in F$ so that by definition of $y$ we have $0 = langle x,y rangle = (n+1)y_n - (n+2)y_{n+1}$, i.e. $$y_{n+1} = frac{n+1}{n+2} , y_n.$$
This is true for all $ngeq 0$. Since we assumed that $y$ has finite support, this is only possible if $y equiv 0$.




As suggested in the comment by Hanno, another way to prove that $F^perp = {0}$ is the following: notice that $phi(x) = langle x , a rangle$ with $a = left( frac{1}{n+1} right)_{n geq 0} in l^2- E$. Then in $l^2$, we have $ker phi = {a}^perp$ and $(ker phi)^perp = text{span}{a}$, so $F^perp = text{span}{a} cap E = {0}$.






share|cite|improve this answer











$endgroup$



Let $E$ be the space of sequences of real numbers with finite support endowed with the $l^2$ inner product and define the linear form $phi colon E to mathbb{R}$ by
$$phi(x) = sum_{n = 0}^infty frac{x_n}{n+1}, quad x in E.$$
Then $phi$ is a continuous linear form, so that $F = ker phi$ is a closed subspace. Notice that $F neq E$ (for instance take $x_0 = 1, x_1=0, x_2 = 0, ldots$). In particular $F$ is not dense in $E$. Finally we can show that $F^perp = {0}$ (see below for a more elegant solution). Indeed let $y in F^perp$ and fix $n geq 0$. Define $x in E$ by
$$begin{align*}x_n &= n+1, \
x_{n+1} &= -(n+2),\
x_k &= 0, quad k notin {n,n+1}.end{align*}$$
Then $x in F$ so that by definition of $y$ we have $0 = langle x,y rangle = (n+1)y_n - (n+2)y_{n+1}$, i.e. $$y_{n+1} = frac{n+1}{n+2} , y_n.$$
This is true for all $ngeq 0$. Since we assumed that $y$ has finite support, this is only possible if $y equiv 0$.




As suggested in the comment by Hanno, another way to prove that $F^perp = {0}$ is the following: notice that $phi(x) = langle x , a rangle$ with $a = left( frac{1}{n+1} right)_{n geq 0} in l^2- E$. Then in $l^2$, we have $ker phi = {a}^perp$ and $(ker phi)^perp = text{span}{a}$, so $F^perp = text{span}{a} cap E = {0}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 11:32

























answered Dec 28 '18 at 2:13









MichhMichh

22316




22316








  • 1




    $begingroup$
    Let me state two points (1) A bit misleading: You write "Notice that $Fneq E$", but the $x$ given inside the parentheses illustrates $kerphine{0}$. (2) Conceptual: The linear form is $phi(x) = langle x,a rangle$ with fixed $a = left(frac 1{n+1}right)_{n=0,1,dots}in ell^2(mathbb N_0)backslash E,$. Then in $ell^2(mathbb N_0)$ one has $(kerphi)^perp = langle arangle$, and $langle aranglecap E ={0}$.
    $endgroup$
    – Hanno
    Dec 29 '18 at 9:55












  • $begingroup$
    @Hanno (1) Yes you are correct about that and I will edit my post. (2) I chose not to talk about $l^2$ because in my eyes it is more confusing than anything else, but again you are correct and it is definitely a more elegant solution. Thanks for the remarks.
    $endgroup$
    – Michh
    Dec 29 '18 at 11:04










  • $begingroup$
    The title of the question would suggest that the OP is looking for an orthonormal subset with this property, not an entire subspace. Can you find an orthonormal basis for $F$? It exists by Gram-Schmidt but that isn't really explicit.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 11:42








  • 1




    $begingroup$
    @mechanodroid Reading off from the answer, the set ${e_n-frac{n+2}{n+1}e_{n+1}mid ninmathbb N_0}$ is a basis for F. Running Gram-Schmidt would yield an orthonormal one, but I do not feel any necessity to carry this through.
    $endgroup$
    – Hanno
    Dec 29 '18 at 12:03










  • $begingroup$
    @Hanno You're right, I agree.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 12:29














  • 1




    $begingroup$
    Let me state two points (1) A bit misleading: You write "Notice that $Fneq E$", but the $x$ given inside the parentheses illustrates $kerphine{0}$. (2) Conceptual: The linear form is $phi(x) = langle x,a rangle$ with fixed $a = left(frac 1{n+1}right)_{n=0,1,dots}in ell^2(mathbb N_0)backslash E,$. Then in $ell^2(mathbb N_0)$ one has $(kerphi)^perp = langle arangle$, and $langle aranglecap E ={0}$.
    $endgroup$
    – Hanno
    Dec 29 '18 at 9:55












  • $begingroup$
    @Hanno (1) Yes you are correct about that and I will edit my post. (2) I chose not to talk about $l^2$ because in my eyes it is more confusing than anything else, but again you are correct and it is definitely a more elegant solution. Thanks for the remarks.
    $endgroup$
    – Michh
    Dec 29 '18 at 11:04










  • $begingroup$
    The title of the question would suggest that the OP is looking for an orthonormal subset with this property, not an entire subspace. Can you find an orthonormal basis for $F$? It exists by Gram-Schmidt but that isn't really explicit.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 11:42








  • 1




    $begingroup$
    @mechanodroid Reading off from the answer, the set ${e_n-frac{n+2}{n+1}e_{n+1}mid ninmathbb N_0}$ is a basis for F. Running Gram-Schmidt would yield an orthonormal one, but I do not feel any necessity to carry this through.
    $endgroup$
    – Hanno
    Dec 29 '18 at 12:03










  • $begingroup$
    @Hanno You're right, I agree.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 12:29








1




1




$begingroup$
Let me state two points (1) A bit misleading: You write "Notice that $Fneq E$", but the $x$ given inside the parentheses illustrates $kerphine{0}$. (2) Conceptual: The linear form is $phi(x) = langle x,a rangle$ with fixed $a = left(frac 1{n+1}right)_{n=0,1,dots}in ell^2(mathbb N_0)backslash E,$. Then in $ell^2(mathbb N_0)$ one has $(kerphi)^perp = langle arangle$, and $langle aranglecap E ={0}$.
$endgroup$
– Hanno
Dec 29 '18 at 9:55






$begingroup$
Let me state two points (1) A bit misleading: You write "Notice that $Fneq E$", but the $x$ given inside the parentheses illustrates $kerphine{0}$. (2) Conceptual: The linear form is $phi(x) = langle x,a rangle$ with fixed $a = left(frac 1{n+1}right)_{n=0,1,dots}in ell^2(mathbb N_0)backslash E,$. Then in $ell^2(mathbb N_0)$ one has $(kerphi)^perp = langle arangle$, and $langle aranglecap E ={0}$.
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– Hanno
Dec 29 '18 at 9:55














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@Hanno (1) Yes you are correct about that and I will edit my post. (2) I chose not to talk about $l^2$ because in my eyes it is more confusing than anything else, but again you are correct and it is definitely a more elegant solution. Thanks for the remarks.
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– Michh
Dec 29 '18 at 11:04




$begingroup$
@Hanno (1) Yes you are correct about that and I will edit my post. (2) I chose not to talk about $l^2$ because in my eyes it is more confusing than anything else, but again you are correct and it is definitely a more elegant solution. Thanks for the remarks.
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– Michh
Dec 29 '18 at 11:04












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The title of the question would suggest that the OP is looking for an orthonormal subset with this property, not an entire subspace. Can you find an orthonormal basis for $F$? It exists by Gram-Schmidt but that isn't really explicit.
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– mechanodroid
Dec 29 '18 at 11:42






$begingroup$
The title of the question would suggest that the OP is looking for an orthonormal subset with this property, not an entire subspace. Can you find an orthonormal basis for $F$? It exists by Gram-Schmidt but that isn't really explicit.
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– mechanodroid
Dec 29 '18 at 11:42






1




1




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@mechanodroid Reading off from the answer, the set ${e_n-frac{n+2}{n+1}e_{n+1}mid ninmathbb N_0}$ is a basis for F. Running Gram-Schmidt would yield an orthonormal one, but I do not feel any necessity to carry this through.
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– Hanno
Dec 29 '18 at 12:03




$begingroup$
@mechanodroid Reading off from the answer, the set ${e_n-frac{n+2}{n+1}e_{n+1}mid ninmathbb N_0}$ is a basis for F. Running Gram-Schmidt would yield an orthonormal one, but I do not feel any necessity to carry this through.
$endgroup$
– Hanno
Dec 29 '18 at 12:03












$begingroup$
@Hanno You're right, I agree.
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– mechanodroid
Dec 29 '18 at 12:29




$begingroup$
@Hanno You're right, I agree.
$endgroup$
– mechanodroid
Dec 29 '18 at 12:29


















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