If $(X,|.|_*)$ is complete and $|.|_{*} leq |.|_{**}$, then is $(X,|.|_{**})$ complete?
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If a space $X$ is complete with respect to $|.|_*$ and we have that $|f|_{*} leq |f|_{**}$ for all $f in X$.
Does this imply that the space will be also complete with respect $|.|_{**}$
functional-analysis normed-spaces complete-spaces
asked Dec 28 '18 at 5:27
Dreamer123Dreamer123
32329
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add a comment |
$begingroup$
If a space $X$ is complete with respect to $|.|_*$ and we have that $|f|_{*} leq |f|_{**}$ for all $f in X$.
Does this imply that the space will be also complete with respect $|.|_{**}$
functional-analysis normed-spaces complete-spaces
asked Dec 28 '18 at 5:27
Dreamer123Dreamer123
32329
$endgroup$
add a comment |
$begingroup$
If a space $X$ is complete with respect to $|.|_*$ and we have that $|f|_{*} leq |f|_{**}$ for all $f in X$.
Does this imply that the space will be also complete with respect $|.|_{**}$
functional-analysis normed-spaces complete-spaces
asked Dec 28 '18 at 5:27
Dreamer123Dreamer123
32329
$endgroup$
If a space $X$ is complete with respect to $|.|_*$ and we have that $|f|_{*} leq |f|_{**}$ for all $f in X$.
Does this imply that the space will be also complete with respect $|.|_{**}$
functional-analysis normed-spaces complete-spaces
functional-analysis normed-spaces complete-spaces
asked Dec 28 '18 at 5:27
Dreamer123Dreamer123
32329
asked Dec 28 '18 at 5:27
Dreamer123Dreamer123
32329
asked Dec 28 '18 at 5:27
Dreamer123Dreamer123
32329
asked Dec 28 '18 at 5:27
Dreamer123Dreamer123
32329
asked Dec 28 '18 at 5:27
Dreamer123Dreamer123
32329
32329
add a comment |
add a comment |
1 Answer
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Using a Hamel basis argument we can show that any infinite dimensional Banach space contains a discontinuous linear functional $f$ such that for some sequence $x_n to 0$ we have $f(x_n) to 1$. Let $|.|_{*}$ be the original norm and $|.|_{**}=|.|_{*}+|f(x)|$. Then the hypothesis is satisfied. But ${x_n}$ is a Cauchy sequence in the second norm which does not converge. [Let ${y_n}$ be linearly independent with $|y_n|=1$ for all $n$. Define $f(y_n)=n$ and extend $f$ linearly to $X$. Then $f$ is not continuous. Also, $frac {y_n} n to 0$ and $f(frac {y_n} n) =1 to 1$. Take $x_n=frac {y_n} n $ in above argument].
answered Dec 28 '18 at 6:11
Kavi Rama MurthyKavi Rama Murthy
61k42262
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1 Answer
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$begingroup$
Using a Hamel basis argument we can show that any infinite dimensional Banach space contains a discontinuous linear functional $f$ such that for some sequence $x_n to 0$ we have $f(x_n) to 1$. Let $|.|_{*}$ be the original norm and $|.|_{**}=|.|_{*}+|f(x)|$. Then the hypothesis is satisfied. But ${x_n}$ is a Cauchy sequence in the second norm which does not converge. [Let ${y_n}$ be linearly independent with $|y_n|=1$ for all $n$. Define $f(y_n)=n$ and extend $f$ linearly to $X$. Then $f$ is not continuous. Also, $frac {y_n} n to 0$ and $f(frac {y_n} n) =1 to 1$. Take $x_n=frac {y_n} n $ in above argument].
answered Dec 28 '18 at 6:11
Kavi Rama MurthyKavi Rama Murthy
61k42262
$endgroup$
add a comment |
$begingroup$
Using a Hamel basis argument we can show that any infinite dimensional Banach space contains a discontinuous linear functional $f$ such that for some sequence $x_n to 0$ we have $f(x_n) to 1$. Let $|.|_{*}$ be the original norm and $|.|_{**}=|.|_{*}+|f(x)|$. Then the hypothesis is satisfied. But ${x_n}$ is a Cauchy sequence in the second norm which does not converge. [Let ${y_n}$ be linearly independent with $|y_n|=1$ for all $n$. Define $f(y_n)=n$ and extend $f$ linearly to $X$. Then $f$ is not continuous. Also, $frac {y_n} n to 0$ and $f(frac {y_n} n) =1 to 1$. Take $x_n=frac {y_n} n $ in above argument].
answered Dec 28 '18 at 6:11
Kavi Rama MurthyKavi Rama Murthy
61k42262
$endgroup$
add a comment |
$begingroup$
Using a Hamel basis argument we can show that any infinite dimensional Banach space contains a discontinuous linear functional $f$ such that for some sequence $x_n to 0$ we have $f(x_n) to 1$. Let $|.|_{*}$ be the original norm and $|.|_{**}=|.|_{*}+|f(x)|$. Then the hypothesis is satisfied. But ${x_n}$ is a Cauchy sequence in the second norm which does not converge. [Let ${y_n}$ be linearly independent with $|y_n|=1$ for all $n$. Define $f(y_n)=n$ and extend $f$ linearly to $X$. Then $f$ is not continuous. Also, $frac {y_n} n to 0$ and $f(frac {y_n} n) =1 to 1$. Take $x_n=frac {y_n} n $ in above argument].
answered Dec 28 '18 at 6:11
Kavi Rama MurthyKavi Rama Murthy
61k42262
$endgroup$
Using a Hamel basis argument we can show that any infinite dimensional Banach space contains a discontinuous linear functional $f$ such that for some sequence $x_n to 0$ we have $f(x_n) to 1$. Let $|.|_{*}$ be the original norm and $|.|_{**}=|.|_{*}+|f(x)|$. Then the hypothesis is satisfied. But ${x_n}$ is a Cauchy sequence in the second norm which does not converge. [Let ${y_n}$ be linearly independent with $|y_n|=1$ for all $n$. Define $f(y_n)=n$ and extend $f$ linearly to $X$. Then $f$ is not continuous. Also, $frac {y_n} n to 0$ and $f(frac {y_n} n) =1 to 1$. Take $x_n=frac {y_n} n $ in above argument].
answered Dec 28 '18 at 6:11
Kavi Rama MurthyKavi Rama Murthy
61k42262
edited Dec 28 '18 at 6:18
answered Dec 28 '18 at 6:11
Kavi Rama MurthyKavi Rama Murthy
61k42262
answered Dec 28 '18 at 6:11
Kavi Rama MurthyKavi Rama Murthy
61k42262
answered Dec 28 '18 at 6:11
Kavi Rama MurthyKavi Rama Murthy
61k42262
61k42262
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