Mixing
Compact open operator between Banach spaces
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Let $X,Y$ be Banach space, $Y$ infinite dimensional. Show that no $T in mathcal{K}(X,Y)$ is open. By definition $T$ is open if and only if $exists r >0$ such that $B_Y(0,r) subset T(B_X(0,1))$ and I also know that the closed unit ball in $Y$ is not compact, since $Y$ is infinite dimensional.
functional-analysis banach-spaces compact-operators open-map
asked Jan 17 at 14:09
user289143user289143
998313
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add a comment |
$begingroup$
Let $X,Y$ be Banach space, $Y$ infinite dimensional. Show that no $T in mathcal{K}(X,Y)$ is open. By definition $T$ is open if and only if $exists r >0$ such that $B_Y(0,r) subset T(B_X(0,1))$ and I also know that the closed unit ball in $Y$ is not compact, since $Y$ is infinite dimensional.
functional-analysis banach-spaces compact-operators open-map
asked Jan 17 at 14:09
user289143user289143
998313
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be Banach space, $Y$ infinite dimensional. Show that no $T in mathcal{K}(X,Y)$ is open. By definition $T$ is open if and only if $exists r >0$ such that $B_Y(0,r) subset T(B_X(0,1))$ and I also know that the closed unit ball in $Y$ is not compact, since $Y$ is infinite dimensional.
functional-analysis banach-spaces compact-operators open-map
asked Jan 17 at 14:09
user289143user289143
998313
$endgroup$
Let $X,Y$ be Banach space, $Y$ infinite dimensional. Show that no $T in mathcal{K}(X,Y)$ is open. By definition $T$ is open if and only if $exists r >0$ such that $B_Y(0,r) subset T(B_X(0,1))$ and I also know that the closed unit ball in $Y$ is not compact, since $Y$ is infinite dimensional.
functional-analysis banach-spaces compact-operators open-map
functional-analysis banach-spaces compact-operators open-map
asked Jan 17 at 14:09
user289143user289143
998313
asked Jan 17 at 14:09
user289143user289143
998313
asked Jan 17 at 14:09
user289143user289143
998313
asked Jan 17 at 14:09
user289143user289143
998313
asked Jan 17 at 14:09
user289143user289143
998313
998313
add a comment |
add a comment |
1 Answer
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$begingroup$
You have all of the right pieces, you just need to assemble them.
Since the closed unit ball is not compact in $Y$, neither is $overline{B}_Y(0,r)$ (since $y mapsto ry$ is a homeomorphism for $r > 0$). Suppose $T$ is open and compact. Then
$$B_Y(0,r) subseteq T(B_X(0,1)) subseteq overline{T(B_X(0,1))}$$
where the last set is compact, since $T$ is a compact operator. This means that $overline{B}_Y(0,r)$ is a closed subset of a compact set and hence is compact which is a contradiction.
answered Jan 17 at 14:18
Rhys SteeleRhys Steele
6,8451829
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$begingroup$
You have all of the right pieces, you just need to assemble them.
Since the closed unit ball is not compact in $Y$, neither is $overline{B}_Y(0,r)$ (since $y mapsto ry$ is a homeomorphism for $r > 0$). Suppose $T$ is open and compact. Then
$$B_Y(0,r) subseteq T(B_X(0,1)) subseteq overline{T(B_X(0,1))}$$
where the last set is compact, since $T$ is a compact operator. This means that $overline{B}_Y(0,r)$ is a closed subset of a compact set and hence is compact which is a contradiction.
answered Jan 17 at 14:18
Rhys SteeleRhys Steele
6,8451829
$endgroup$
add a comment |
$begingroup$
You have all of the right pieces, you just need to assemble them.
Since the closed unit ball is not compact in $Y$, neither is $overline{B}_Y(0,r)$ (since $y mapsto ry$ is a homeomorphism for $r > 0$). Suppose $T$ is open and compact. Then
$$B_Y(0,r) subseteq T(B_X(0,1)) subseteq overline{T(B_X(0,1))}$$
where the last set is compact, since $T$ is a compact operator. This means that $overline{B}_Y(0,r)$ is a closed subset of a compact set and hence is compact which is a contradiction.
answered Jan 17 at 14:18
Rhys SteeleRhys Steele
6,8451829
$endgroup$
add a comment |
$begingroup$
You have all of the right pieces, you just need to assemble them.
Since the closed unit ball is not compact in $Y$, neither is $overline{B}_Y(0,r)$ (since $y mapsto ry$ is a homeomorphism for $r > 0$). Suppose $T$ is open and compact. Then
$$B_Y(0,r) subseteq T(B_X(0,1)) subseteq overline{T(B_X(0,1))}$$
where the last set is compact, since $T$ is a compact operator. This means that $overline{B}_Y(0,r)$ is a closed subset of a compact set and hence is compact which is a contradiction.
answered Jan 17 at 14:18
Rhys SteeleRhys Steele
6,8451829
$endgroup$
You have all of the right pieces, you just need to assemble them.
Since the closed unit ball is not compact in $Y$, neither is $overline{B}_Y(0,r)$ (since $y mapsto ry$ is a homeomorphism for $r > 0$). Suppose $T$ is open and compact. Then
$$B_Y(0,r) subseteq T(B_X(0,1)) subseteq overline{T(B_X(0,1))}$$
where the last set is compact, since $T$ is a compact operator. This means that $overline{B}_Y(0,r)$ is a closed subset of a compact set and hence is compact which is a contradiction.
answered Jan 17 at 14:18
Rhys SteeleRhys Steele
6,8451829
answered Jan 17 at 14:18
Rhys SteeleRhys Steele
6,8451829
answered Jan 17 at 14:18
Rhys SteeleRhys Steele
6,8451829
answered Jan 17 at 14:18
Rhys SteeleRhys Steele
6,8451829
6,8451829
add a comment |
add a comment |
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