I want to show explicitly that $L^2[0,1]$ subspace of $L^1[0,1]$ has no internal points
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I want to show explicitly that $L^2[0,1]$ subspace of $L^1[0,1]$ has no internal points.
If S subspace of X, I have the following result:
S = X if and only if the interior of S $neq emptyset$
Then to prove that the interior is empty, I just need to prove S is different from X.
Can I just pick $f(x) = frac{1}{sqrt{x}}$ and say that it belongs to $L^1$ but not to $L^2$ and be done?
Is there something that I am missing?
vector-spaces lp-spaces
asked Nov 19 at 21:52
qcc101
458113
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I want to show explicitly that $L^2[0,1]$ subspace of $L^1[0,1]$ has no internal points.
If S subspace of X, I have the following result:
S = X if and only if the interior of S $neq emptyset$
Then to prove that the interior is empty, I just need to prove S is different from X.
Can I just pick $f(x) = frac{1}{sqrt{x}}$ and say that it belongs to $L^1$ but not to $L^2$ and be done?
Is there something that I am missing?
vector-spaces lp-spaces
asked Nov 19 at 21:52
qcc101
458113
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to show explicitly that $L^2[0,1]$ subspace of $L^1[0,1]$ has no internal points.
If S subspace of X, I have the following result:
S = X if and only if the interior of S $neq emptyset$
Then to prove that the interior is empty, I just need to prove S is different from X.
Can I just pick $f(x) = frac{1}{sqrt{x}}$ and say that it belongs to $L^1$ but not to $L^2$ and be done?
Is there something that I am missing?
vector-spaces lp-spaces
asked Nov 19 at 21:52
qcc101
458113
I want to show explicitly that $L^2[0,1]$ subspace of $L^1[0,1]$ has no internal points.
If S subspace of X, I have the following result:
S = X if and only if the interior of S $neq emptyset$
Then to prove that the interior is empty, I just need to prove S is different from X.
Can I just pick $f(x) = frac{1}{sqrt{x}}$ and say that it belongs to $L^1$ but not to $L^2$ and be done?
Is there something that I am missing?
vector-spaces lp-spaces
vector-spaces lp-spaces
asked Nov 19 at 21:52
qcc101
458113
asked Nov 19 at 21:52
qcc101
458113
asked Nov 19 at 21:52
qcc101
458113
asked Nov 19 at 21:52
qcc101
458113
asked Nov 19 at 21:52
qcc101
458113
458113
add a comment |
add a comment |
2 Answers
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3
down vote
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No, you are not missing anything. If $X,Y$ are normed spaces, and $X$ is a proper subspace of $Y$, then $X$ has empty interior relative to $Y$. So if you show that $L^2[0,1]$ is a proper subspace of $L^1[0,1]$, you are done. The function $f(x)=frac{1}{sqrt{x}}$ is indeed an element of $L^1[0,1]$ that does not belong to $L^2[0,1]$.
answered Nov 19 at 22:01
uniquesolution
8,631823
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I think 'explicitly' here means you have to show that if $f in L^{2}$ then $B(f,epsilon)$ is not contained in $L^{1}$ for any $epsilon >0$. General theory is not supposed to be used here. For this just consider $f+frac 1 {nsqrt x}$ and show that this function lies in the ball $B(f,epsilon)$ for $n$ sufficiently large but it does not belong to $L^{2}$ for any $n$.
answered Nov 19 at 23:48
Kavi Rama Murthy
46.9k31854
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
No, you are not missing anything. If $X,Y$ are normed spaces, and $X$ is a proper subspace of $Y$, then $X$ has empty interior relative to $Y$. So if you show that $L^2[0,1]$ is a proper subspace of $L^1[0,1]$, you are done. The function $f(x)=frac{1}{sqrt{x}}$ is indeed an element of $L^1[0,1]$ that does not belong to $L^2[0,1]$.
answered Nov 19 at 22:01
uniquesolution
8,631823
add a comment |
up vote
3
down vote
accepted
No, you are not missing anything. If $X,Y$ are normed spaces, and $X$ is a proper subspace of $Y$, then $X$ has empty interior relative to $Y$. So if you show that $L^2[0,1]$ is a proper subspace of $L^1[0,1]$, you are done. The function $f(x)=frac{1}{sqrt{x}}$ is indeed an element of $L^1[0,1]$ that does not belong to $L^2[0,1]$.
answered Nov 19 at 22:01
uniquesolution
8,631823
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
No, you are not missing anything. If $X,Y$ are normed spaces, and $X$ is a proper subspace of $Y$, then $X$ has empty interior relative to $Y$. So if you show that $L^2[0,1]$ is a proper subspace of $L^1[0,1]$, you are done. The function $f(x)=frac{1}{sqrt{x}}$ is indeed an element of $L^1[0,1]$ that does not belong to $L^2[0,1]$.
answered Nov 19 at 22:01
uniquesolution
8,631823
No, you are not missing anything. If $X,Y$ are normed spaces, and $X$ is a proper subspace of $Y$, then $X$ has empty interior relative to $Y$. So if you show that $L^2[0,1]$ is a proper subspace of $L^1[0,1]$, you are done. The function $f(x)=frac{1}{sqrt{x}}$ is indeed an element of $L^1[0,1]$ that does not belong to $L^2[0,1]$.
answered Nov 19 at 22:01
uniquesolution
8,631823
answered Nov 19 at 22:01
uniquesolution
8,631823
answered Nov 19 at 22:01
uniquesolution
8,631823
answered Nov 19 at 22:01
uniquesolution
8,631823
8,631823
add a comment |
add a comment |
up vote
1
down vote
I think 'explicitly' here means you have to show that if $f in L^{2}$ then $B(f,epsilon)$ is not contained in $L^{1}$ for any $epsilon >0$. General theory is not supposed to be used here. For this just consider $f+frac 1 {nsqrt x}$ and show that this function lies in the ball $B(f,epsilon)$ for $n$ sufficiently large but it does not belong to $L^{2}$ for any $n$.
answered Nov 19 at 23:48
Kavi Rama Murthy
46.9k31854
add a comment |
up vote
1
down vote
I think 'explicitly' here means you have to show that if $f in L^{2}$ then $B(f,epsilon)$ is not contained in $L^{1}$ for any $epsilon >0$. General theory is not supposed to be used here. For this just consider $f+frac 1 {nsqrt x}$ and show that this function lies in the ball $B(f,epsilon)$ for $n$ sufficiently large but it does not belong to $L^{2}$ for any $n$.
answered Nov 19 at 23:48
Kavi Rama Murthy
46.9k31854
add a comment |
up vote
1
down vote
up vote
1
down vote
I think 'explicitly' here means you have to show that if $f in L^{2}$ then $B(f,epsilon)$ is not contained in $L^{1}$ for any $epsilon >0$. General theory is not supposed to be used here. For this just consider $f+frac 1 {nsqrt x}$ and show that this function lies in the ball $B(f,epsilon)$ for $n$ sufficiently large but it does not belong to $L^{2}$ for any $n$.
answered Nov 19 at 23:48
Kavi Rama Murthy
46.9k31854
I think 'explicitly' here means you have to show that if $f in L^{2}$ then $B(f,epsilon)$ is not contained in $L^{1}$ for any $epsilon >0$. General theory is not supposed to be used here. For this just consider $f+frac 1 {nsqrt x}$ and show that this function lies in the ball $B(f,epsilon)$ for $n$ sufficiently large but it does not belong to $L^{2}$ for any $n$.
answered Nov 19 at 23:48
Kavi Rama Murthy
46.9k31854
answered Nov 19 at 23:48
Kavi Rama Murthy
46.9k31854
answered Nov 19 at 23:48
Kavi Rama Murthy
46.9k31854
answered Nov 19 at 23:48
Kavi Rama Murthy
46.9k31854
46.9k31854
add a comment |
add a comment |
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