Banach Space，pde，compact operator
$begingroup$
$X, Y, Z$ are Banach Space, $f: X to Y$, $g: Y to Z$ are bounded linear operators. Show: if $f$ or $g$ is a compact operator , then $g circ f: X to Z$ is a compact operator.
functional-analysis
edited Jan 10 at 4:46
staedtlerr
1587
asked Jan 10 at 3:30
user628743user628743
84
$endgroup$
add a comment |
$begingroup$
$X, Y, Z$ are Banach Space, $f: X to Y$, $g: Y to Z$ are bounded linear operators. Show: if $f$ or $g$ is a compact operator , then $g circ f: X to Z$ is a compact operator.
functional-analysis
edited Jan 10 at 4:46
staedtlerr
1587
asked Jan 10 at 3:30
user628743user628743
84
$endgroup$
add a comment |
$begingroup$
$X, Y, Z$ are Banach Space, $f: X to Y$, $g: Y to Z$ are bounded linear operators. Show: if $f$ or $g$ is a compact operator , then $g circ f: X to Z$ is a compact operator.
functional-analysis
edited Jan 10 at 4:46
staedtlerr
1587
asked Jan 10 at 3:30
user628743user628743
84
$endgroup$
$X, Y, Z$ are Banach Space, $f: X to Y$, $g: Y to Z$ are bounded linear operators. Show: if $f$ or $g$ is a compact operator , then $g circ f: X to Z$ is a compact operator.
functional-analysis
functional-analysis
edited Jan 10 at 4:46
staedtlerr
1587
asked Jan 10 at 3:30
user628743user628743
84
edited Jan 10 at 4:46
staedtlerr
1587
asked Jan 10 at 3:30
user628743user628743
84
edited Jan 10 at 4:46
staedtlerr
1587
edited Jan 10 at 4:46
staedtlerr
1587
edited Jan 10 at 4:46
staedtlerr
1587
1587
asked Jan 10 at 3:30
user628743user628743
84
asked Jan 10 at 3:30
user628743user628743
84
asked Jan 10 at 3:30
user628743user628743
84
84
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Suppose first that $g$ is compact, and let $x_i$ be a bounded sequence in $X$; then $f(x_i)$ is a bounded sequence in $Y$, since $f$ is a bounded operator; then the compactness of $g$ implies that $g circ f(x_i) = g (f(x_i))$ has a covergent subsequence; hence $g circ f$ is compact.
Likewise if $f$ is compact, the sequence $f(x_i)$ has a convergent subsequence; thus so does $g circ f(x_i) = g(f(x_i))$ by the continuity of $g$, which is equivalent to its boundedness. Thus $f circ g$ is compact in this case as well.
answered Jan 10 at 3:44
Robert LewisRobert Lewis
48.7k23167
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add a comment |
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$begingroup$
Suppose first that $g$ is compact, and let $x_i$ be a bounded sequence in $X$; then $f(x_i)$ is a bounded sequence in $Y$, since $f$ is a bounded operator; then the compactness of $g$ implies that $g circ f(x_i) = g (f(x_i))$ has a covergent subsequence; hence $g circ f$ is compact.
Likewise if $f$ is compact, the sequence $f(x_i)$ has a convergent subsequence; thus so does $g circ f(x_i) = g(f(x_i))$ by the continuity of $g$, which is equivalent to its boundedness. Thus $f circ g$ is compact in this case as well.
answered Jan 10 at 3:44
Robert LewisRobert Lewis
48.7k23167
$endgroup$
add a comment |
$begingroup$
Suppose first that $g$ is compact, and let $x_i$ be a bounded sequence in $X$; then $f(x_i)$ is a bounded sequence in $Y$, since $f$ is a bounded operator; then the compactness of $g$ implies that $g circ f(x_i) = g (f(x_i))$ has a covergent subsequence; hence $g circ f$ is compact.
Likewise if $f$ is compact, the sequence $f(x_i)$ has a convergent subsequence; thus so does $g circ f(x_i) = g(f(x_i))$ by the continuity of $g$, which is equivalent to its boundedness. Thus $f circ g$ is compact in this case as well.
answered Jan 10 at 3:44
Robert LewisRobert Lewis
48.7k23167
$endgroup$
add a comment |
$begingroup$
Suppose first that $g$ is compact, and let $x_i$ be a bounded sequence in $X$; then $f(x_i)$ is a bounded sequence in $Y$, since $f$ is a bounded operator; then the compactness of $g$ implies that $g circ f(x_i) = g (f(x_i))$ has a covergent subsequence; hence $g circ f$ is compact.
Likewise if $f$ is compact, the sequence $f(x_i)$ has a convergent subsequence; thus so does $g circ f(x_i) = g(f(x_i))$ by the continuity of $g$, which is equivalent to its boundedness. Thus $f circ g$ is compact in this case as well.
answered Jan 10 at 3:44
Robert LewisRobert Lewis
48.7k23167
$endgroup$
Suppose first that $g$ is compact, and let $x_i$ be a bounded sequence in $X$; then $f(x_i)$ is a bounded sequence in $Y$, since $f$ is a bounded operator; then the compactness of $g$ implies that $g circ f(x_i) = g (f(x_i))$ has a covergent subsequence; hence $g circ f$ is compact.
Likewise if $f$ is compact, the sequence $f(x_i)$ has a convergent subsequence; thus so does $g circ f(x_i) = g(f(x_i))$ by the continuity of $g$, which is equivalent to its boundedness. Thus $f circ g$ is compact in this case as well.
answered Jan 10 at 3:44
Robert LewisRobert Lewis
48.7k23167
answered Jan 10 at 3:44
Robert LewisRobert Lewis
48.7k23167
answered Jan 10 at 3:44
Robert LewisRobert Lewis
48.7k23167
answered Jan 10 at 3:44
Robert LewisRobert Lewis
48.7k23167
48.7k23167
add a comment |
add a comment |
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