Mixing
Closed subset of polynomials in function space
$begingroup$
Not for homework, I am trying to self study functional analysis and encountered the following problem.
We let $C[0,1/2]$ the continous functions defined on that subset of the real line. We look at a subspace of $C[0,1/2]$, consisting of all polynomials on $[0,1/2]$ and call it $W$. Given $delta>0$, set.
$g(x)=deltasum_{n=1}^{infty}frac{1}{n+1}x^n$
Where $xin[0,1/2]$. First, we are tasked with showing that $g$ is in the open ball $B(0,delta)$. I suppose this is done by showing that the norm of $g$ must be less than delta (we are using the supremum norm inherited from $C[0,1/2]$). But from that result, we are tasked with using it to conclude that $W$ cannot be an open subset of $C[0,1/2]$. So I must show that for some $win W$, there is no $epsilon$, such that an open ball $B(v,epsilon)$ is in $W$? Not sure how that follows from the previous?
real-analysis functional-analysis convergence vector-spaces normed-spaces
asked Jan 2 at 22:56
thaumoctopusthaumoctopus
1
$endgroup$
add a comment |
$begingroup$
Not for homework, I am trying to self study functional analysis and encountered the following problem.
We let $C[0,1/2]$ the continous functions defined on that subset of the real line. We look at a subspace of $C[0,1/2]$, consisting of all polynomials on $[0,1/2]$ and call it $W$. Given $delta>0$, set.
$g(x)=deltasum_{n=1}^{infty}frac{1}{n+1}x^n$
Where $xin[0,1/2]$. First, we are tasked with showing that $g$ is in the open ball $B(0,delta)$. I suppose this is done by showing that the norm of $g$ must be less than delta (we are using the supremum norm inherited from $C[0,1/2]$). But from that result, we are tasked with using it to conclude that $W$ cannot be an open subset of $C[0,1/2]$. So I must show that for some $win W$, there is no $epsilon$, such that an open ball $B(v,epsilon)$ is in $W$? Not sure how that follows from the previous?
real-analysis functional-analysis convergence vector-spaces normed-spaces
asked Jan 2 at 22:56
thaumoctopusthaumoctopus
1
$endgroup$
1
$begingroup$
To answer your first question, yes, you need to show the norm of $g$ is less than $delta$. As for your second question, note that if you define $g_k(x) = delta sum_{n=1}^k x^n/(n + 1)$ then $g_k in W$. If $W$ is closed, do you know anything about how the limit of the sequence $(g_k)_{k geq 1}$ must behave? (This is just a characterisation of closedness different to the one you have provided.)
$endgroup$
– Riley
Jan 2 at 23:13
1
$begingroup$
A proper subspace of a normed space is never open. In fact, it even has empty interior.
$endgroup$
– mechanodroid
Jan 3 at 12:16
add a comment |
$begingroup$
Not for homework, I am trying to self study functional analysis and encountered the following problem.
We let $C[0,1/2]$ the continous functions defined on that subset of the real line. We look at a subspace of $C[0,1/2]$, consisting of all polynomials on $[0,1/2]$ and call it $W$. Given $delta>0$, set.
$g(x)=deltasum_{n=1}^{infty}frac{1}{n+1}x^n$
Where $xin[0,1/2]$. First, we are tasked with showing that $g$ is in the open ball $B(0,delta)$. I suppose this is done by showing that the norm of $g$ must be less than delta (we are using the supremum norm inherited from $C[0,1/2]$). But from that result, we are tasked with using it to conclude that $W$ cannot be an open subset of $C[0,1/2]$. So I must show that for some $win W$, there is no $epsilon$, such that an open ball $B(v,epsilon)$ is in $W$? Not sure how that follows from the previous?
real-analysis functional-analysis convergence vector-spaces normed-spaces
asked Jan 2 at 22:56
thaumoctopusthaumoctopus
1
$endgroup$
Not for homework, I am trying to self study functional analysis and encountered the following problem.
We let $C[0,1/2]$ the continous functions defined on that subset of the real line. We look at a subspace of $C[0,1/2]$, consisting of all polynomials on $[0,1/2]$ and call it $W$. Given $delta>0$, set.
$g(x)=deltasum_{n=1}^{infty}frac{1}{n+1}x^n$
Where $xin[0,1/2]$. First, we are tasked with showing that $g$ is in the open ball $B(0,delta)$. I suppose this is done by showing that the norm of $g$ must be less than delta (we are using the supremum norm inherited from $C[0,1/2]$). But from that result, we are tasked with using it to conclude that $W$ cannot be an open subset of $C[0,1/2]$. So I must show that for some $win W$, there is no $epsilon$, such that an open ball $B(v,epsilon)$ is in $W$? Not sure how that follows from the previous?
real-analysis functional-analysis convergence vector-spaces normed-spaces
real-analysis functional-analysis convergence vector-spaces normed-spaces
asked Jan 2 at 22:56
thaumoctopusthaumoctopus
1
asked Jan 2 at 22:56
thaumoctopusthaumoctopus
1
asked Jan 2 at 22:56
thaumoctopusthaumoctopus
1
asked Jan 2 at 22:56
thaumoctopusthaumoctopus
1
asked Jan 2 at 22:56
thaumoctopusthaumoctopus
1
1
1
$begingroup$
To answer your first question, yes, you need to show the norm of $g$ is less than $delta$. As for your second question, note that if you define $g_k(x) = delta sum_{n=1}^k x^n/(n + 1)$ then $g_k in W$. If $W$ is closed, do you know anything about how the limit of the sequence $(g_k)_{k geq 1}$ must behave? (This is just a characterisation of closedness different to the one you have provided.)
$endgroup$
– Riley
Jan 2 at 23:13
1
$begingroup$
A proper subspace of a normed space is never open. In fact, it even has empty interior.
$endgroup$
– mechanodroid
Jan 3 at 12:16
add a comment |
1
$begingroup$
To answer your first question, yes, you need to show the norm of $g$ is less than $delta$. As for your second question, note that if you define $g_k(x) = delta sum_{n=1}^k x^n/(n + 1)$ then $g_k in W$. If $W$ is closed, do you know anything about how the limit of the sequence $(g_k)_{k geq 1}$ must behave? (This is just a characterisation of closedness different to the one you have provided.)
$endgroup$
– Riley
Jan 2 at 23:13
1
$begingroup$
A proper subspace of a normed space is never open. In fact, it even has empty interior.
$endgroup$
– mechanodroid
Jan 3 at 12:16
1
1
$begingroup$
To answer your first question, yes, you need to show the norm of $g$ is less than $delta$. As for your second question, note that if you define $g_k(x) = delta sum_{n=1}^k x^n/(n + 1)$ then $g_k in W$. If $W$ is closed, do you know anything about how the limit of the sequence $(g_k)_{k geq 1}$ must behave? (This is just a characterisation of closedness different to the one you have provided.)
$endgroup$
– Riley
Jan 2 at 23:13
$begingroup$
To answer your first question, yes, you need to show the norm of $g$ is less than $delta$. As for your second question, note that if you define $g_k(x) = delta sum_{n=1}^k x^n/(n + 1)$ then $g_k in W$. If $W$ is closed, do you know anything about how the limit of the sequence $(g_k)_{k geq 1}$ must behave? (This is just a characterisation of closedness different to the one you have provided.)
$endgroup$
– Riley
Jan 2 at 23:13
1
1
$begingroup$
A proper subspace of a normed space is never open. In fact, it even has empty interior.
$endgroup$
– mechanodroid
Jan 3 at 12:16
$begingroup$
A proper subspace of a normed space is never open. In fact, it even has empty interior.
$endgroup$
– mechanodroid
Jan 3 at 12:16
add a comment |
2 Answers
2
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$begingroup$
$|g| < delta sum_{n=1}^{infty} frac 1 {2^{n}}= delta$. This proves the first part. Now,suppose $W$ is open . Since the zero polynomial is in $W$ there must exist $delta >0$ such that $B(0,delta) subset W$. Consider the $g$ corresponding to this $delta$. Then $g in B(0,delta)$ so we must have $g in W$. Can you see that this is a contradiction? [It is a known fact that if $sum a_n x^{n}$ converges for $|x| leq r$ and if the sum is zero for all such $x$ the $a_n=0$ for all $n$].
answered Jan 2 at 23:14
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
add a comment |
$begingroup$
(1). Suppose that $Y$ is a vector subspace of a normed linear space $X$ and that $Y$ has non-empty interior. Then $ Y=X:$ For some $r>0$ we have $B(0,r)subset Y,$ but then ( because $Y$ is a vector space), $Ysupset cup_{nin Bbb N}{nv: vin B(0,r)}=cup_{nin Bbb N}B(0,nr)=X.$
The reason $B(0,r)subset Y$ for some $r>0$ is that for some $yin Y$ and some $r>0$ we have $B(y,r)subset Y,$ and since $Y$ is a vector space we have $Ysupset {y'-y:y'in B(y,r)}=B(0,r). $
(2). $W$ is a vector subspace of $C[0,1/2],$ so to prove that $W$ has empty interior in $C[0,1/2]$, it suffices to show $Wne C[0,1/2].$ Let $f(t)=|t-1/4|$
for $tin [0,1/2]$. Then $fin C[0,1/2],$ but $f$ is not a polynomial because $f(t)$ is not differentiable at $t=1/4$.
answered Jan 3 at 19:22
DanielWainfleetDanielWainfleet
34.6k31648
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
$|g| < delta sum_{n=1}^{infty} frac 1 {2^{n}}= delta$. This proves the first part. Now,suppose $W$ is open . Since the zero polynomial is in $W$ there must exist $delta >0$ such that $B(0,delta) subset W$. Consider the $g$ corresponding to this $delta$. Then $g in B(0,delta)$ so we must have $g in W$. Can you see that this is a contradiction? [It is a known fact that if $sum a_n x^{n}$ converges for $|x| leq r$ and if the sum is zero for all such $x$ the $a_n=0$ for all $n$].
answered Jan 2 at 23:14
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
add a comment |
$begingroup$
$|g| < delta sum_{n=1}^{infty} frac 1 {2^{n}}= delta$. This proves the first part. Now,suppose $W$ is open . Since the zero polynomial is in $W$ there must exist $delta >0$ such that $B(0,delta) subset W$. Consider the $g$ corresponding to this $delta$. Then $g in B(0,delta)$ so we must have $g in W$. Can you see that this is a contradiction? [It is a known fact that if $sum a_n x^{n}$ converges for $|x| leq r$ and if the sum is zero for all such $x$ the $a_n=0$ for all $n$].
answered Jan 2 at 23:14
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
add a comment |
$begingroup$
$|g| < delta sum_{n=1}^{infty} frac 1 {2^{n}}= delta$. This proves the first part. Now,suppose $W$ is open . Since the zero polynomial is in $W$ there must exist $delta >0$ such that $B(0,delta) subset W$. Consider the $g$ corresponding to this $delta$. Then $g in B(0,delta)$ so we must have $g in W$. Can you see that this is a contradiction? [It is a known fact that if $sum a_n x^{n}$ converges for $|x| leq r$ and if the sum is zero for all such $x$ the $a_n=0$ for all $n$].
answered Jan 2 at 23:14
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
$|g| < delta sum_{n=1}^{infty} frac 1 {2^{n}}= delta$. This proves the first part. Now,suppose $W$ is open . Since the zero polynomial is in $W$ there must exist $delta >0$ such that $B(0,delta) subset W$. Consider the $g$ corresponding to this $delta$. Then $g in B(0,delta)$ so we must have $g in W$. Can you see that this is a contradiction? [It is a known fact that if $sum a_n x^{n}$ converges for $|x| leq r$ and if the sum is zero for all such $x$ the $a_n=0$ for all $n$].
answered Jan 2 at 23:14
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
edited Jan 2 at 23:19
answered Jan 2 at 23:14
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
answered Jan 2 at 23:14
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
answered Jan 2 at 23:14
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
53.5k32055
add a comment |
add a comment |
$begingroup$
(1). Suppose that $Y$ is a vector subspace of a normed linear space $X$ and that $Y$ has non-empty interior. Then $ Y=X:$ For some $r>0$ we have $B(0,r)subset Y,$ but then ( because $Y$ is a vector space), $Ysupset cup_{nin Bbb N}{nv: vin B(0,r)}=cup_{nin Bbb N}B(0,nr)=X.$
The reason $B(0,r)subset Y$ for some $r>0$ is that for some $yin Y$ and some $r>0$ we have $B(y,r)subset Y,$ and since $Y$ is a vector space we have $Ysupset {y'-y:y'in B(y,r)}=B(0,r). $
(2). $W$ is a vector subspace of $C[0,1/2],$ so to prove that $W$ has empty interior in $C[0,1/2]$, it suffices to show $Wne C[0,1/2].$ Let $f(t)=|t-1/4|$
for $tin [0,1/2]$. Then $fin C[0,1/2],$ but $f$ is not a polynomial because $f(t)$ is not differentiable at $t=1/4$.
answered Jan 3 at 19:22
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
add a comment |
$begingroup$
(1). Suppose that $Y$ is a vector subspace of a normed linear space $X$ and that $Y$ has non-empty interior. Then $ Y=X:$ For some $r>0$ we have $B(0,r)subset Y,$ but then ( because $Y$ is a vector space), $Ysupset cup_{nin Bbb N}{nv: vin B(0,r)}=cup_{nin Bbb N}B(0,nr)=X.$
The reason $B(0,r)subset Y$ for some $r>0$ is that for some $yin Y$ and some $r>0$ we have $B(y,r)subset Y,$ and since $Y$ is a vector space we have $Ysupset {y'-y:y'in B(y,r)}=B(0,r). $
(2). $W$ is a vector subspace of $C[0,1/2],$ so to prove that $W$ has empty interior in $C[0,1/2]$, it suffices to show $Wne C[0,1/2].$ Let $f(t)=|t-1/4|$
for $tin [0,1/2]$. Then $fin C[0,1/2],$ but $f$ is not a polynomial because $f(t)$ is not differentiable at $t=1/4$.
answered Jan 3 at 19:22
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
add a comment |
$begingroup$
(1). Suppose that $Y$ is a vector subspace of a normed linear space $X$ and that $Y$ has non-empty interior. Then $ Y=X:$ For some $r>0$ we have $B(0,r)subset Y,$ but then ( because $Y$ is a vector space), $Ysupset cup_{nin Bbb N}{nv: vin B(0,r)}=cup_{nin Bbb N}B(0,nr)=X.$
The reason $B(0,r)subset Y$ for some $r>0$ is that for some $yin Y$ and some $r>0$ we have $B(y,r)subset Y,$ and since $Y$ is a vector space we have $Ysupset {y'-y:y'in B(y,r)}=B(0,r). $
(2). $W$ is a vector subspace of $C[0,1/2],$ so to prove that $W$ has empty interior in $C[0,1/2]$, it suffices to show $Wne C[0,1/2].$ Let $f(t)=|t-1/4|$
for $tin [0,1/2]$. Then $fin C[0,1/2],$ but $f$ is not a polynomial because $f(t)$ is not differentiable at $t=1/4$.
answered Jan 3 at 19:22
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
(1). Suppose that $Y$ is a vector subspace of a normed linear space $X$ and that $Y$ has non-empty interior. Then $ Y=X:$ For some $r>0$ we have $B(0,r)subset Y,$ but then ( because $Y$ is a vector space), $Ysupset cup_{nin Bbb N}{nv: vin B(0,r)}=cup_{nin Bbb N}B(0,nr)=X.$
The reason $B(0,r)subset Y$ for some $r>0$ is that for some $yin Y$ and some $r>0$ we have $B(y,r)subset Y,$ and since $Y$ is a vector space we have $Ysupset {y'-y:y'in B(y,r)}=B(0,r). $
(2). $W$ is a vector subspace of $C[0,1/2],$ so to prove that $W$ has empty interior in $C[0,1/2]$, it suffices to show $Wne C[0,1/2].$ Let $f(t)=|t-1/4|$
for $tin [0,1/2]$. Then $fin C[0,1/2],$ but $f$ is not a polynomial because $f(t)$ is not differentiable at $t=1/4$.
answered Jan 3 at 19:22
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 19:22
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 19:22
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 19:22
DanielWainfleetDanielWainfleet
34.6k31648
34.6k31648
add a comment |
add a comment |
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$begingroup$
To answer your first question, yes, you need to show the norm of $g$ is less than $delta$. As for your second question, note that if you define $g_k(x) = delta sum_{n=1}^k x^n/(n + 1)$ then $g_k in W$. If $W$ is closed, do you know anything about how the limit of the sequence $(g_k)_{k geq 1}$ must behave? (This is just a characterisation of closedness different to the one you have provided.)
$endgroup$
– Riley
Jan 2 at 23:13
1
$begingroup$
A proper subspace of a normed space is never open. In fact, it even has empty interior.
$endgroup$
– mechanodroid
Jan 3 at 12:16