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If E* is separable, so is E. Is this proof correct?
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This is my first time posting here. I already saw a post regarding this result, but I wanted to check if this specific proof is correct or not; as the proof presented in the book is different. Also, I'm not a native English speaker, so excuse me if there's any grammar mistakes.
Proposition: Let $(E,||.||)$ be a normed space (over $mathbbR$ or $mathbbC$) such that $E^*$ is separable. Then $E$ is separable as well.
Proof: Let $D = phi_n_n in mathbbN$ be a dense in $E^*$.
Define $x_n in E$ such that $||x_n||=1$ and $||phi_n|| < |phi_n(x_n)|+frac1n$, which exists by definition of $||phi_n||$. I claim that $langle x_1,...x_k...rangle$ is a dense in $E$. If I could prove that, due to a previous result, this would imply that $E$ is separable. We'll define $S=overlinelangle x_1,...x_k...rangle$. We need to prove that $S=E$.
Let's assume otherwise. Then there would exist $x in E-S$, and given that $S$ is a subspace of $E$ we can assume $||x||=1$. By a corollary of the Hahn-Banach theorem (since $S$ is closed), we can find a continious functional $phi in E^*$ such that $phi(S)=0$ and $phi(x)=1$.
Now, we know that there exists a sequence in $D$ such that $phi_n_k rightarrow phi$, but if that's the case:
$$||phi_n_k||-frac1n_k<|phi_n_k(x_n_k)|=|phi_n_k(x_n_k)-0|=|phi_n_k(x_n_k)-phi(x_n_k)|leq ||phi_n_k-phi||,$$
which implies that $||phi_n_k|| rightarrow 0$ as $k rightarrow infty$. For $k$ large enough, $|phi_n_k(y)| < frac12$ for all $y$ with $||y||=1$, in particular, $|phi_n_k(x)| < frac12$. Then,
$$||phi-phi_n_k|| geq |phi(x)-phi_n_k(x)| = |1-phi_n_k(x)| >frac12$$
But $||phi-phi_n_k|| rightarrow 0$? We reached our contradiction and conclude that $S=E$.
Thank you in advance.
functional-analysis normed-spaces dual-spaces
asked Aug 3 at 4:38
PedroR
212
add a comment |Â
up vote
4
down vote
favorite
This is my first time posting here. I already saw a post regarding this result, but I wanted to check if this specific proof is correct or not; as the proof presented in the book is different. Also, I'm not a native English speaker, so excuse me if there's any grammar mistakes.
Proposition: Let $(E,||.||)$ be a normed space (over $mathbbR$ or $mathbbC$) such that $E^*$ is separable. Then $E$ is separable as well.
Proof: Let $D = phi_n_n in mathbbN$ be a dense in $E^*$.
Define $x_n in E$ such that $||x_n||=1$ and $||phi_n|| < |phi_n(x_n)|+frac1n$, which exists by definition of $||phi_n||$. I claim that $langle x_1,...x_k...rangle$ is a dense in $E$. If I could prove that, due to a previous result, this would imply that $E$ is separable. We'll define $S=overlinelangle x_1,...x_k...rangle$. We need to prove that $S=E$.
Let's assume otherwise. Then there would exist $x in E-S$, and given that $S$ is a subspace of $E$ we can assume $||x||=1$. By a corollary of the Hahn-Banach theorem (since $S$ is closed), we can find a continious functional $phi in E^*$ such that $phi(S)=0$ and $phi(x)=1$.
Now, we know that there exists a sequence in $D$ such that $phi_n_k rightarrow phi$, but if that's the case:
$$||phi_n_k||-frac1n_k<|phi_n_k(x_n_k)|=|phi_n_k(x_n_k)-0|=|phi_n_k(x_n_k)-phi(x_n_k)|leq ||phi_n_k-phi||,$$
which implies that $||phi_n_k|| rightarrow 0$ as $k rightarrow infty$. For $k$ large enough, $|phi_n_k(y)| < frac12$ for all $y$ with $||y||=1$, in particular, $|phi_n_k(x)| < frac12$. Then,
$$||phi-phi_n_k|| geq |phi(x)-phi_n_k(x)| = |1-phi_n_k(x)| >frac12$$
But $||phi-phi_n_k|| rightarrow 0$? We reached our contradiction and conclude that $S=E$.
Thank you in advance.
functional-analysis normed-spaces dual-spaces
asked Aug 3 at 4:38
PedroR
212
1
Looks great, maybe toward the end, as soon as you showed $|phi_n_k| rightarrow 0$ you can conclude the contradiction since $|phi|geq phi(x) = 1$ with reverse triangle inequality.
– Xiao
Aug 3 at 4:56
Oh, yeah, that's a nicer way to put it. I could also say that since $||.||$ is continious, $phi = 0$. Thanks!
– PedroR
Aug 3 at 5:03
2
There's also a typo. When $|phi(x_n_k) - 0|$ is written, I think you mean $|phi_n_k(x_n_k) - 0|$.
– Theo Bendit
Aug 3 at 5:06
Fixed. Thank you!
– PedroR
Aug 3 at 5:09
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This is my first time posting here. I already saw a post regarding this result, but I wanted to check if this specific proof is correct or not; as the proof presented in the book is different. Also, I'm not a native English speaker, so excuse me if there's any grammar mistakes.
Proposition: Let $(E,||.||)$ be a normed space (over $mathbbR$ or $mathbbC$) such that $E^*$ is separable. Then $E$ is separable as well.
Proof: Let $D = phi_n_n in mathbbN$ be a dense in $E^*$.
Define $x_n in E$ such that $||x_n||=1$ and $||phi_n|| < |phi_n(x_n)|+frac1n$, which exists by definition of $||phi_n||$. I claim that $langle x_1,...x_k...rangle$ is a dense in $E$. If I could prove that, due to a previous result, this would imply that $E$ is separable. We'll define $S=overlinelangle x_1,...x_k...rangle$. We need to prove that $S=E$.
Let's assume otherwise. Then there would exist $x in E-S$, and given that $S$ is a subspace of $E$ we can assume $||x||=1$. By a corollary of the Hahn-Banach theorem (since $S$ is closed), we can find a continious functional $phi in E^*$ such that $phi(S)=0$ and $phi(x)=1$.
Now, we know that there exists a sequence in $D$ such that $phi_n_k rightarrow phi$, but if that's the case:
$$||phi_n_k||-frac1n_k<|phi_n_k(x_n_k)|=|phi_n_k(x_n_k)-0|=|phi_n_k(x_n_k)-phi(x_n_k)|leq ||phi_n_k-phi||,$$
which implies that $||phi_n_k|| rightarrow 0$ as $k rightarrow infty$. For $k$ large enough, $|phi_n_k(y)| < frac12$ for all $y$ with $||y||=1$, in particular, $|phi_n_k(x)| < frac12$. Then,
$$||phi-phi_n_k|| geq |phi(x)-phi_n_k(x)| = |1-phi_n_k(x)| >frac12$$
But $||phi-phi_n_k|| rightarrow 0$? We reached our contradiction and conclude that $S=E$.
Thank you in advance.
functional-analysis normed-spaces dual-spaces
asked Aug 3 at 4:38
PedroR
212
This is my first time posting here. I already saw a post regarding this result, but I wanted to check if this specific proof is correct or not; as the proof presented in the book is different. Also, I'm not a native English speaker, so excuse me if there's any grammar mistakes.
Proposition: Let $(E,||.||)$ be a normed space (over $mathbbR$ or $mathbbC$) such that $E^*$ is separable. Then $E$ is separable as well.
Proof: Let $D = phi_n_n in mathbbN$ be a dense in $E^*$.
Define $x_n in E$ such that $||x_n||=1$ and $||phi_n|| < |phi_n(x_n)|+frac1n$, which exists by definition of $||phi_n||$. I claim that $langle x_1,...x_k...rangle$ is a dense in $E$. If I could prove that, due to a previous result, this would imply that $E$ is separable. We'll define $S=overlinelangle x_1,...x_k...rangle$. We need to prove that $S=E$.
Let's assume otherwise. Then there would exist $x in E-S$, and given that $S$ is a subspace of $E$ we can assume $||x||=1$. By a corollary of the Hahn-Banach theorem (since $S$ is closed), we can find a continious functional $phi in E^*$ such that $phi(S)=0$ and $phi(x)=1$.
Now, we know that there exists a sequence in $D$ such that $phi_n_k rightarrow phi$, but if that's the case:
$$||phi_n_k||-frac1n_k<|phi_n_k(x_n_k)|=|phi_n_k(x_n_k)-0|=|phi_n_k(x_n_k)-phi(x_n_k)|leq ||phi_n_k-phi||,$$
which implies that $||phi_n_k|| rightarrow 0$ as $k rightarrow infty$. For $k$ large enough, $|phi_n_k(y)| < frac12$ for all $y$ with $||y||=1$, in particular, $|phi_n_k(x)| < frac12$. Then,
$$||phi-phi_n_k|| geq |phi(x)-phi_n_k(x)| = |1-phi_n_k(x)| >frac12$$
But $||phi-phi_n_k|| rightarrow 0$? We reached our contradiction and conclude that $S=E$.
Thank you in advance.
functional-analysis normed-spaces dual-spaces
asked Aug 3 at 4:38
PedroR
212
edited Aug 3 at 5:09
asked Aug 3 at 4:38
PedroR
212
asked Aug 3 at 4:38
PedroR
212
asked Aug 3 at 4:38
PedroR
212
212
1
Looks great, maybe toward the end, as soon as you showed $|phi_n_k| rightarrow 0$ you can conclude the contradiction since $|phi|geq phi(x) = 1$ with reverse triangle inequality.
– Xiao
Aug 3 at 4:56
Oh, yeah, that's a nicer way to put it. I could also say that since $||.||$ is continious, $phi = 0$. Thanks!
– PedroR
Aug 3 at 5:03
2
There's also a typo. When $|phi(x_n_k) - 0|$ is written, I think you mean $|phi_n_k(x_n_k) - 0|$.
– Theo Bendit
Aug 3 at 5:06
Fixed. Thank you!
– PedroR
Aug 3 at 5:09
add a comment |Â
1
Looks great, maybe toward the end, as soon as you showed $|phi_n_k| rightarrow 0$ you can conclude the contradiction since $|phi|geq phi(x) = 1$ with reverse triangle inequality.
– Xiao
Aug 3 at 4:56
Oh, yeah, that's a nicer way to put it. I could also say that since $||.||$ is continious, $phi = 0$. Thanks!
– PedroR
Aug 3 at 5:03
2
There's also a typo. When $|phi(x_n_k) - 0|$ is written, I think you mean $|phi_n_k(x_n_k) - 0|$.
– Theo Bendit
Aug 3 at 5:06
Fixed. Thank you!
– PedroR
Aug 3 at 5:09
1
1
Looks great, maybe toward the end, as soon as you showed $|phi_n_k| rightarrow 0$ you can conclude the contradiction since $|phi|geq phi(x) = 1$ with reverse triangle inequality.
– Xiao
Aug 3 at 4:56
Looks great, maybe toward the end, as soon as you showed $|phi_n_k| rightarrow 0$ you can conclude the contradiction since $|phi|geq phi(x) = 1$ with reverse triangle inequality.
– Xiao
Aug 3 at 4:56
Oh, yeah, that's a nicer way to put it. I could also say that since $||.||$ is continious, $phi = 0$. Thanks!
– PedroR
Aug 3 at 5:03
Oh, yeah, that's a nicer way to put it. I could also say that since $||.||$ is continious, $phi = 0$. Thanks!
– PedroR
Aug 3 at 5:03
2
2
There's also a typo. When $|phi(x_n_k) - 0|$ is written, I think you mean $|phi_n_k(x_n_k) - 0|$.
– Theo Bendit
Aug 3 at 5:06
There's also a typo. When $|phi(x_n_k) - 0|$ is written, I think you mean $|phi_n_k(x_n_k) - 0|$.
– Theo Bendit
Aug 3 at 5:06
Fixed. Thank you!
– PedroR
Aug 3 at 5:09
Fixed. Thank you!
– PedroR
Aug 3 at 5:09
add a comment |Â
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1
Looks great, maybe toward the end, as soon as you showed $|phi_n_k| rightarrow 0$ you can conclude the contradiction since $|phi|geq phi(x) = 1$ with reverse triangle inequality.
– Xiao
Aug 3 at 4:56
Oh, yeah, that's a nicer way to put it. I could also say that since $||.||$ is continious, $phi = 0$. Thanks!
– PedroR
Aug 3 at 5:03
2
There's also a typo. When $|phi(x_n_k) - 0|$ is written, I think you mean $|phi_n_k(x_n_k) - 0|$.
– Theo Bendit
Aug 3 at 5:06
Fixed. Thank you!
– PedroR
Aug 3 at 5:09