If $T:Xto Y$ is a continuous linear map at the origin, then $T$ is Lipschitz
$begingroup$
Let $X$ and $Y$ be normed linear spaces over a scalar field, and let $T:Xto Y$ be a linear map.
Suppose $T:Xto Y$ is a continuous at the origin, I want to show that $T$ is Lipschitz, i.e. there exists some constant $Kgeq 0,$ such that begin{align} Vert T(x) Vert leq KVert xVert ,;;forall;xin X. end{align}
HERE IS A PROOF
Suppose for contradiction, that $forall, nin Bbb{N},exists,x_nin X$ such $x_nneq 0$ and begin{align} Vert T(x_n) Vert > nVert x_nVert . end{align}
So, begin{align} frac{Vert T(x_n) Vert}{nVert x_nVert} > 1 ,;;forall, nin Bbb{N}. end{align}
Define begin{align} u_n=frac{x_n}{nVert x_nVert} ,;;forall, nin Bbb{N}. end{align}
Then, begin{align} u_nto 0, ;;text{as} ;;ntoinfty, end{align}
but begin{align} Vert T(u_n) -0Vert=frac{Vert T(x_n) Vert}{nVert x_nVert} > 1 ,;;forall, nin Bbb{N}. end{align}
This implies that $T(u_n)notto 0, ;;text{as} ;;ntoinfty.$ Contadiction and we're done.
My question: Why must $x_nneq 0 ,;;forall, nin Bbb{N}$ at negation?
functional-analysis normed-spaces
asked Dec 17 '18 at 9:38
Omojola MichealOmojola Micheal
1,969324
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be normed linear spaces over a scalar field, and let $T:Xto Y$ be a linear map.
Suppose $T:Xto Y$ is a continuous at the origin, I want to show that $T$ is Lipschitz, i.e. there exists some constant $Kgeq 0,$ such that begin{align} Vert T(x) Vert leq KVert xVert ,;;forall;xin X. end{align}
HERE IS A PROOF
Suppose for contradiction, that $forall, nin Bbb{N},exists,x_nin X$ such $x_nneq 0$ and begin{align} Vert T(x_n) Vert > nVert x_nVert . end{align}
So, begin{align} frac{Vert T(x_n) Vert}{nVert x_nVert} > 1 ,;;forall, nin Bbb{N}. end{align}
Define begin{align} u_n=frac{x_n}{nVert x_nVert} ,;;forall, nin Bbb{N}. end{align}
Then, begin{align} u_nto 0, ;;text{as} ;;ntoinfty, end{align}
but begin{align} Vert T(u_n) -0Vert=frac{Vert T(x_n) Vert}{nVert x_nVert} > 1 ,;;forall, nin Bbb{N}. end{align}
This implies that $T(u_n)notto 0, ;;text{as} ;;ntoinfty.$ Contadiction and we're done.
My question: Why must $x_nneq 0 ,;;forall, nin Bbb{N}$ at negation?
functional-analysis normed-spaces
asked Dec 17 '18 at 9:38
Omojola MichealOmojola Micheal
1,969324
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be normed linear spaces over a scalar field, and let $T:Xto Y$ be a linear map.
Suppose $T:Xto Y$ is a continuous at the origin, I want to show that $T$ is Lipschitz, i.e. there exists some constant $Kgeq 0,$ such that begin{align} Vert T(x) Vert leq KVert xVert ,;;forall;xin X. end{align}
HERE IS A PROOF
Suppose for contradiction, that $forall, nin Bbb{N},exists,x_nin X$ such $x_nneq 0$ and begin{align} Vert T(x_n) Vert > nVert x_nVert . end{align}
So, begin{align} frac{Vert T(x_n) Vert}{nVert x_nVert} > 1 ,;;forall, nin Bbb{N}. end{align}
Define begin{align} u_n=frac{x_n}{nVert x_nVert} ,;;forall, nin Bbb{N}. end{align}
Then, begin{align} u_nto 0, ;;text{as} ;;ntoinfty, end{align}
but begin{align} Vert T(u_n) -0Vert=frac{Vert T(x_n) Vert}{nVert x_nVert} > 1 ,;;forall, nin Bbb{N}. end{align}
This implies that $T(u_n)notto 0, ;;text{as} ;;ntoinfty.$ Contadiction and we're done.
My question: Why must $x_nneq 0 ,;;forall, nin Bbb{N}$ at negation?
functional-analysis normed-spaces
asked Dec 17 '18 at 9:38
Omojola MichealOmojola Micheal
1,969324
$endgroup$
Let $X$ and $Y$ be normed linear spaces over a scalar field, and let $T:Xto Y$ be a linear map.
Suppose $T:Xto Y$ is a continuous at the origin, I want to show that $T$ is Lipschitz, i.e. there exists some constant $Kgeq 0,$ such that begin{align} Vert T(x) Vert leq KVert xVert ,;;forall;xin X. end{align}
HERE IS A PROOF
Suppose for contradiction, that $forall, nin Bbb{N},exists,x_nin X$ such $x_nneq 0$ and begin{align} Vert T(x_n) Vert > nVert x_nVert . end{align}
So, begin{align} frac{Vert T(x_n) Vert}{nVert x_nVert} > 1 ,;;forall, nin Bbb{N}. end{align}
Define begin{align} u_n=frac{x_n}{nVert x_nVert} ,;;forall, nin Bbb{N}. end{align}
Then, begin{align} u_nto 0, ;;text{as} ;;ntoinfty, end{align}
but begin{align} Vert T(u_n) -0Vert=frac{Vert T(x_n) Vert}{nVert x_nVert} > 1 ,;;forall, nin Bbb{N}. end{align}
This implies that $T(u_n)notto 0, ;;text{as} ;;ntoinfty.$ Contadiction and we're done.
My question: Why must $x_nneq 0 ,;;forall, nin Bbb{N}$ at negation?
functional-analysis normed-spaces
functional-analysis normed-spaces
asked Dec 17 '18 at 9:38
Omojola MichealOmojola Micheal
1,969324
asked Dec 17 '18 at 9:38
Omojola MichealOmojola Micheal
1,969324
asked Dec 17 '18 at 9:38
Omojola MichealOmojola Micheal
1,969324
asked Dec 17 '18 at 9:38
Omojola MichealOmojola Micheal
1,969324
asked Dec 17 '18 at 9:38
Omojola MichealOmojola Micheal
1,969324
1,969324
add a comment |
add a comment |
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If you chose $x_n$ such that $|Tx_n| >n|x_n|$ and $x_n=0$ you get $0>0$! So $x_n neq 0$ automatically.
answered Dec 17 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
$endgroup$
$begingroup$
Oh, I see Sir! Thanks! +1)
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:41
$begingroup$
I believe I can still go directly and use the argument. Since $T(x_n)to 0,$ then it is bounded. This implies that it is Lipschitz. Right?
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:43
add a comment |
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$begingroup$
If you chose $x_n$ such that $|Tx_n| >n|x_n|$ and $x_n=0$ you get $0>0$! So $x_n neq 0$ automatically.
answered Dec 17 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
$endgroup$
$begingroup$
Oh, I see Sir! Thanks! +1)
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:41
$begingroup$
I believe I can still go directly and use the argument. Since $T(x_n)to 0,$ then it is bounded. This implies that it is Lipschitz. Right?
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:43
add a comment |
$begingroup$
If you chose $x_n$ such that $|Tx_n| >n|x_n|$ and $x_n=0$ you get $0>0$! So $x_n neq 0$ automatically.
answered Dec 17 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
$endgroup$
$begingroup$
Oh, I see Sir! Thanks! +1)
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:41
$begingroup$
I believe I can still go directly and use the argument. Since $T(x_n)to 0,$ then it is bounded. This implies that it is Lipschitz. Right?
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:43
add a comment |
$begingroup$
If you chose $x_n$ such that $|Tx_n| >n|x_n|$ and $x_n=0$ you get $0>0$! So $x_n neq 0$ automatically.
answered Dec 17 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
$endgroup$
If you chose $x_n$ such that $|Tx_n| >n|x_n|$ and $x_n=0$ you get $0>0$! So $x_n neq 0$ automatically.
answered Dec 17 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
answered Dec 17 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
answered Dec 17 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
answered Dec 17 '18 at 9:40
Kavi Rama MurthyKavi Rama Murthy
66.7k52867
66.7k52867
$begingroup$
Oh, I see Sir! Thanks! +1)
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:41
$begingroup$
I believe I can still go directly and use the argument. Since $T(x_n)to 0,$ then it is bounded. This implies that it is Lipschitz. Right?
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:43
add a comment |
$begingroup$
Oh, I see Sir! Thanks! +1)
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:41
$begingroup$
I believe I can still go directly and use the argument. Since $T(x_n)to 0,$ then it is bounded. This implies that it is Lipschitz. Right?
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:43
$begingroup$
Oh, I see Sir! Thanks! +1)
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:41
$begingroup$
Oh, I see Sir! Thanks! +1)
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:41
$begingroup$
I believe I can still go directly and use the argument. Since $T(x_n)to 0,$ then it is bounded. This implies that it is Lipschitz. Right?
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:43
$begingroup$
I believe I can still go directly and use the argument. Since $T(x_n)to 0,$ then it is bounded. This implies that it is Lipschitz. Right?
$endgroup$
– Omojola Micheal
Dec 17 '18 at 9:43
add a comment |
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