Triangle inequality for linear operators
$begingroup$
For a linear operator $R in B(X,Y)$ and some $k in mathbb{R}_{>0}$, do we have
$$(forall x in X:lVert R(x) rVert leq k lVert x rVert) Rightarrow lVert R rVert leq k
$$?
I'd like to use this fact to prove the triangle inequality for norms of linear operators $S, T in B(X,Y)$, of which I know that
$$forall x in X:lVert (S+T)(x)rVert leq (lVert S rVert + rVert T lVert) lVert xrVert
$$
analysis operator-theory norm
asked Jan 17 at 18:25
Jos van NieuwmanJos van Nieuwman
439
$endgroup$
add a comment |
$begingroup$
For a linear operator $R in B(X,Y)$ and some $k in mathbb{R}_{>0}$, do we have
$$(forall x in X:lVert R(x) rVert leq k lVert x rVert) Rightarrow lVert R rVert leq k
$$?
I'd like to use this fact to prove the triangle inequality for norms of linear operators $S, T in B(X,Y)$, of which I know that
$$forall x in X:lVert (S+T)(x)rVert leq (lVert S rVert + rVert T lVert) lVert xrVert
$$
analysis operator-theory norm
asked Jan 17 at 18:25
Jos van NieuwmanJos van Nieuwman
439
$endgroup$
add a comment |
$begingroup$
For a linear operator $R in B(X,Y)$ and some $k in mathbb{R}_{>0}$, do we have
$$(forall x in X:lVert R(x) rVert leq k lVert x rVert) Rightarrow lVert R rVert leq k
$$?
I'd like to use this fact to prove the triangle inequality for norms of linear operators $S, T in B(X,Y)$, of which I know that
$$forall x in X:lVert (S+T)(x)rVert leq (lVert S rVert + rVert T lVert) lVert xrVert
$$
analysis operator-theory norm
asked Jan 17 at 18:25
Jos van NieuwmanJos van Nieuwman
439
$endgroup$
For a linear operator $R in B(X,Y)$ and some $k in mathbb{R}_{>0}$, do we have
$$(forall x in X:lVert R(x) rVert leq k lVert x rVert) Rightarrow lVert R rVert leq k
$$?
I'd like to use this fact to prove the triangle inequality for norms of linear operators $S, T in B(X,Y)$, of which I know that
$$forall x in X:lVert (S+T)(x)rVert leq (lVert S rVert + rVert T lVert) lVert xrVert
$$
analysis operator-theory norm
analysis operator-theory norm
asked Jan 17 at 18:25
Jos van NieuwmanJos van Nieuwman
439
asked Jan 17 at 18:25
Jos van NieuwmanJos van Nieuwman
439
asked Jan 17 at 18:25
Jos van NieuwmanJos van Nieuwman
439
asked Jan 17 at 18:25
Jos van NieuwmanJos van Nieuwman
439
asked Jan 17 at 18:25
Jos van NieuwmanJos van Nieuwman
439
439
add a comment |
add a comment |
1 Answer
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$begingroup$
Note that
$$|R|=sup{|R(x)|:xin X,|x|=1}.$$
Indeed, write $p(R)=sup{|R(x)|:xin X,|x|=1}$. Since
$${|R(x)|:xin X,|x|=1}subset{|R(x)|:xin X,|x|leq1},$$
we have $p(R)leq|R|$. On the other hand, if $0<|x|leq1$, then
$$|R(x)|leqfrac{1}{|x|}|R(x)|=left|Rleft(frac{x}{|x|}right)right|leq p(R).$$
Taking the supremum over all such $x$ yields $|R|leq p(R)$, and thus $|R|=p(R)$.
If now $xin X$ and $|x|=1$, then by assumption $|R(x)|leq k|x|=k$. Thus $k$ is an upper bound of ${|R(x)|:xin X,|x|=1}$, whence $|R|leq k$.
answered Jan 17 at 18:28
AweyganAweygan
14.2k21441
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Note that
$$|R|=sup{|R(x)|:xin X,|x|=1}.$$
Indeed, write $p(R)=sup{|R(x)|:xin X,|x|=1}$. Since
$${|R(x)|:xin X,|x|=1}subset{|R(x)|:xin X,|x|leq1},$$
we have $p(R)leq|R|$. On the other hand, if $0<|x|leq1$, then
$$|R(x)|leqfrac{1}{|x|}|R(x)|=left|Rleft(frac{x}{|x|}right)right|leq p(R).$$
Taking the supremum over all such $x$ yields $|R|leq p(R)$, and thus $|R|=p(R)$.
If now $xin X$ and $|x|=1$, then by assumption $|R(x)|leq k|x|=k$. Thus $k$ is an upper bound of ${|R(x)|:xin X,|x|=1}$, whence $|R|leq k$.
answered Jan 17 at 18:28
AweyganAweygan
14.2k21441
$endgroup$
add a comment |
$begingroup$
Note that
$$|R|=sup{|R(x)|:xin X,|x|=1}.$$
Indeed, write $p(R)=sup{|R(x)|:xin X,|x|=1}$. Since
$${|R(x)|:xin X,|x|=1}subset{|R(x)|:xin X,|x|leq1},$$
we have $p(R)leq|R|$. On the other hand, if $0<|x|leq1$, then
$$|R(x)|leqfrac{1}{|x|}|R(x)|=left|Rleft(frac{x}{|x|}right)right|leq p(R).$$
Taking the supremum over all such $x$ yields $|R|leq p(R)$, and thus $|R|=p(R)$.
If now $xin X$ and $|x|=1$, then by assumption $|R(x)|leq k|x|=k$. Thus $k$ is an upper bound of ${|R(x)|:xin X,|x|=1}$, whence $|R|leq k$.
answered Jan 17 at 18:28
AweyganAweygan
14.2k21441
$endgroup$
add a comment |
$begingroup$
Note that
$$|R|=sup{|R(x)|:xin X,|x|=1}.$$
Indeed, write $p(R)=sup{|R(x)|:xin X,|x|=1}$. Since
$${|R(x)|:xin X,|x|=1}subset{|R(x)|:xin X,|x|leq1},$$
we have $p(R)leq|R|$. On the other hand, if $0<|x|leq1$, then
$$|R(x)|leqfrac{1}{|x|}|R(x)|=left|Rleft(frac{x}{|x|}right)right|leq p(R).$$
Taking the supremum over all such $x$ yields $|R|leq p(R)$, and thus $|R|=p(R)$.
If now $xin X$ and $|x|=1$, then by assumption $|R(x)|leq k|x|=k$. Thus $k$ is an upper bound of ${|R(x)|:xin X,|x|=1}$, whence $|R|leq k$.
answered Jan 17 at 18:28
AweyganAweygan
14.2k21441
$endgroup$
Note that
$$|R|=sup{|R(x)|:xin X,|x|=1}.$$
Indeed, write $p(R)=sup{|R(x)|:xin X,|x|=1}$. Since
$${|R(x)|:xin X,|x|=1}subset{|R(x)|:xin X,|x|leq1},$$
we have $p(R)leq|R|$. On the other hand, if $0<|x|leq1$, then
$$|R(x)|leqfrac{1}{|x|}|R(x)|=left|Rleft(frac{x}{|x|}right)right|leq p(R).$$
Taking the supremum over all such $x$ yields $|R|leq p(R)$, and thus $|R|=p(R)$.
If now $xin X$ and $|x|=1$, then by assumption $|R(x)|leq k|x|=k$. Thus $k$ is an upper bound of ${|R(x)|:xin X,|x|=1}$, whence $|R|leq k$.
answered Jan 17 at 18:28
AweyganAweygan
14.2k21441
edited Jan 17 at 18:51
answered Jan 17 at 18:28
AweyganAweygan
14.2k21441
answered Jan 17 at 18:28
AweyganAweygan
14.2k21441
answered Jan 17 at 18:28
AweyganAweygan
14.2k21441
14.2k21441
add a comment |
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