Operator norm $ ( ell_2 to ell_1)$
$begingroup$
Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. $ T:X→Y$.
I want to calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.
I do not know how to continue
$$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq 0} frac{∥(x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq 0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $$
functional-analysis operator-theory norm normed-spaces lp-spaces
asked Dec 16 '18 at 20:16
Anna SchmitzAnna Schmitz
917
$endgroup$
add a comment |
$begingroup$
Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. $ T:X→Y$.
I want to calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.
I do not know how to continue
$$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq 0} frac{∥(x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq 0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $$
functional-analysis operator-theory norm normed-spaces lp-spaces
asked Dec 16 '18 at 20:16
Anna SchmitzAnna Schmitz
917
$endgroup$
$begingroup$
I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
$endgroup$
– hyperkahler
Dec 16 '18 at 20:30
$begingroup$
I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
$endgroup$
– Anna Schmitz
Dec 16 '18 at 20:38
add a comment |
$begingroup$
Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. $ T:X→Y$.
I want to calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.
I do not know how to continue
$$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq 0} frac{∥(x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq 0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $$
functional-analysis operator-theory norm normed-spaces lp-spaces
asked Dec 16 '18 at 20:16
Anna SchmitzAnna Schmitz
917
$endgroup$
Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. $ T:X→Y$.
I want to calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.
I do not know how to continue
$$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq 0} frac{∥(x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq 0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $$
functional-analysis operator-theory norm normed-spaces lp-spaces
functional-analysis operator-theory norm normed-spaces lp-spaces
asked Dec 16 '18 at 20:16
Anna SchmitzAnna Schmitz
917
asked Dec 16 '18 at 20:16
Anna SchmitzAnna Schmitz
917
edited Dec 18 '18 at 15:34
user593746
asked Dec 16 '18 at 20:16
Anna SchmitzAnna Schmitz
917
asked Dec 16 '18 at 20:16
Anna SchmitzAnna Schmitz
917
asked Dec 16 '18 at 20:16
Anna SchmitzAnna Schmitz
917
917
$begingroup$
I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
$endgroup$
– hyperkahler
Dec 16 '18 at 20:30
$begingroup$
I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
$endgroup$
– Anna Schmitz
Dec 16 '18 at 20:38
add a comment |
$begingroup$
I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
$endgroup$
– hyperkahler
Dec 16 '18 at 20:30
$begingroup$
I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
$endgroup$
– Anna Schmitz
Dec 16 '18 at 20:38
$begingroup$
I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
$endgroup$
– hyperkahler
Dec 16 '18 at 20:30
$begingroup$
I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
$endgroup$
– hyperkahler
Dec 16 '18 at 20:30
$begingroup$
I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
$endgroup$
– Anna Schmitz
Dec 16 '18 at 20:38
$begingroup$
I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
$endgroup$
– Anna Schmitz
Dec 16 '18 at 20:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I will elaborate on my comment above
Given an operator $T: X rightarrow Y$ where $$X = mathbb{K}^{n}$$
$$Y = l^{1}(mathbb{N})$$ its norm is given by
$$||T||_{op} = sup_{x neq 0}{frac{||Tx||_{1}}{||x||_{2}}} = sup_{||x||_{2} leq 1}{||Tx||_{1}} = sup_{||x||_{2} = 1}{||Tx||_{1}}$$
So we have to maximize $$||Tx||_{1} = |x_{1}| + ldots + |x_{n}|$$ given $$||x||_{2} := (x_{1}^{2} + ldots + x_{n}^{2})^{frac{1}{2}} = 1$$
Let $t_{i} := |x_{i}|$, then our problem reformulates as follows:
$$t_{1} + t_{2} + ldots + t_{n} rightarrow text{max}$$
$$t_{1}^{2} + t_{2}^{2} + ldots + t_{n}^{2} = 1$$
$$t_{i} geq 0, forall i = 1, ldots, n$$
The Lagrangian for the problem is:
$$L = (t_{1} + ldots + t_{n}) - lambda (t_{1}^{2} + ldots + t_{n}^{2} - 1)$$
and $lambda$ stays for the lagrange multiplier.
The necessary extremum condition implies:
$$frac{partial L}{partial t_{i}} := 1 - lambda(2t_{i}) = 0$$
thus $t_{i} = frac{1}{2 lambda}$. Since $t_{i} geq 0$ it follows that $lambda > 0$.
The stationary point is $x = (frac{1}{2 lambda}, ldots, frac{1}{2 lambda})$ and we shall find the lambda such that the solution satisfy the second condition in the problem above.
$||x||_{2} = 1$ is equivalent to
$$ n cdot frac{1}{4 lambda^{2}} = 1$$ from which we get
$lambda = frac{sqrt{n}}{2}$
(note that we shall choose it positive due to the restrictions mentioned above)
Thus $t_{i} = frac{1}{sqrt{n}}$ and the maximum equals $$text{max} = n cdot frac{1}{sqrt{n}} = sqrt{n}$$
answered Dec 16 '18 at 20:52
hyperkahlerhyperkahler
1,483714
$endgroup$
1
$begingroup$
Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
$endgroup$
– Rebellos
Dec 16 '18 at 21:16
$begingroup$
@Rebellos i've made a mistake, please, check the answer for the updates
$endgroup$
– hyperkahler
Dec 16 '18 at 21:25
add a comment |
$begingroup$
Using Cauchy-Schwarz inequality, we have
$$
sum_{i=1}^nleftlvert x_irightrvert=sum_{i=1}^nleftlvert x_irightrvertcdot 1leqslant left(sum_{i=1}^nx_i^2right)^{1/2}left(sum_{i=1}^n1right)^{1/2}=sqrt nleft(sum_{i=1}^nx_i^2right)^{1/2}
$$
hence $leftlVert TrightrVertleqslant sqrt n$.
For the opposite inequality, look at the case where $x_i=1$ for all $iin{1,dots,n}$.
answered Dec 18 '18 at 10:56
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I will elaborate on my comment above
Given an operator $T: X rightarrow Y$ where $$X = mathbb{K}^{n}$$
$$Y = l^{1}(mathbb{N})$$ its norm is given by
$$||T||_{op} = sup_{x neq 0}{frac{||Tx||_{1}}{||x||_{2}}} = sup_{||x||_{2} leq 1}{||Tx||_{1}} = sup_{||x||_{2} = 1}{||Tx||_{1}}$$
So we have to maximize $$||Tx||_{1} = |x_{1}| + ldots + |x_{n}|$$ given $$||x||_{2} := (x_{1}^{2} + ldots + x_{n}^{2})^{frac{1}{2}} = 1$$
Let $t_{i} := |x_{i}|$, then our problem reformulates as follows:
$$t_{1} + t_{2} + ldots + t_{n} rightarrow text{max}$$
$$t_{1}^{2} + t_{2}^{2} + ldots + t_{n}^{2} = 1$$
$$t_{i} geq 0, forall i = 1, ldots, n$$
The Lagrangian for the problem is:
$$L = (t_{1} + ldots + t_{n}) - lambda (t_{1}^{2} + ldots + t_{n}^{2} - 1)$$
and $lambda$ stays for the lagrange multiplier.
The necessary extremum condition implies:
$$frac{partial L}{partial t_{i}} := 1 - lambda(2t_{i}) = 0$$
thus $t_{i} = frac{1}{2 lambda}$. Since $t_{i} geq 0$ it follows that $lambda > 0$.
The stationary point is $x = (frac{1}{2 lambda}, ldots, frac{1}{2 lambda})$ and we shall find the lambda such that the solution satisfy the second condition in the problem above.
$||x||_{2} = 1$ is equivalent to
$$ n cdot frac{1}{4 lambda^{2}} = 1$$ from which we get
$lambda = frac{sqrt{n}}{2}$
(note that we shall choose it positive due to the restrictions mentioned above)
Thus $t_{i} = frac{1}{sqrt{n}}$ and the maximum equals $$text{max} = n cdot frac{1}{sqrt{n}} = sqrt{n}$$
answered Dec 16 '18 at 20:52
hyperkahlerhyperkahler
1,483714
$endgroup$
1
$begingroup$
Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
$endgroup$
– Rebellos
Dec 16 '18 at 21:16
$begingroup$
@Rebellos i've made a mistake, please, check the answer for the updates
$endgroup$
– hyperkahler
Dec 16 '18 at 21:25
add a comment |
$begingroup$
I will elaborate on my comment above
Given an operator $T: X rightarrow Y$ where $$X = mathbb{K}^{n}$$
$$Y = l^{1}(mathbb{N})$$ its norm is given by
$$||T||_{op} = sup_{x neq 0}{frac{||Tx||_{1}}{||x||_{2}}} = sup_{||x||_{2} leq 1}{||Tx||_{1}} = sup_{||x||_{2} = 1}{||Tx||_{1}}$$
So we have to maximize $$||Tx||_{1} = |x_{1}| + ldots + |x_{n}|$$ given $$||x||_{2} := (x_{1}^{2} + ldots + x_{n}^{2})^{frac{1}{2}} = 1$$
Let $t_{i} := |x_{i}|$, then our problem reformulates as follows:
$$t_{1} + t_{2} + ldots + t_{n} rightarrow text{max}$$
$$t_{1}^{2} + t_{2}^{2} + ldots + t_{n}^{2} = 1$$
$$t_{i} geq 0, forall i = 1, ldots, n$$
The Lagrangian for the problem is:
$$L = (t_{1} + ldots + t_{n}) - lambda (t_{1}^{2} + ldots + t_{n}^{2} - 1)$$
and $lambda$ stays for the lagrange multiplier.
The necessary extremum condition implies:
$$frac{partial L}{partial t_{i}} := 1 - lambda(2t_{i}) = 0$$
thus $t_{i} = frac{1}{2 lambda}$. Since $t_{i} geq 0$ it follows that $lambda > 0$.
The stationary point is $x = (frac{1}{2 lambda}, ldots, frac{1}{2 lambda})$ and we shall find the lambda such that the solution satisfy the second condition in the problem above.
$||x||_{2} = 1$ is equivalent to
$$ n cdot frac{1}{4 lambda^{2}} = 1$$ from which we get
$lambda = frac{sqrt{n}}{2}$
(note that we shall choose it positive due to the restrictions mentioned above)
Thus $t_{i} = frac{1}{sqrt{n}}$ and the maximum equals $$text{max} = n cdot frac{1}{sqrt{n}} = sqrt{n}$$
answered Dec 16 '18 at 20:52
hyperkahlerhyperkahler
1,483714
$endgroup$
1
$begingroup$
Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
$endgroup$
– Rebellos
Dec 16 '18 at 21:16
$begingroup$
@Rebellos i've made a mistake, please, check the answer for the updates
$endgroup$
– hyperkahler
Dec 16 '18 at 21:25
add a comment |
$begingroup$
I will elaborate on my comment above
Given an operator $T: X rightarrow Y$ where $$X = mathbb{K}^{n}$$
$$Y = l^{1}(mathbb{N})$$ its norm is given by
$$||T||_{op} = sup_{x neq 0}{frac{||Tx||_{1}}{||x||_{2}}} = sup_{||x||_{2} leq 1}{||Tx||_{1}} = sup_{||x||_{2} = 1}{||Tx||_{1}}$$
So we have to maximize $$||Tx||_{1} = |x_{1}| + ldots + |x_{n}|$$ given $$||x||_{2} := (x_{1}^{2} + ldots + x_{n}^{2})^{frac{1}{2}} = 1$$
Let $t_{i} := |x_{i}|$, then our problem reformulates as follows:
$$t_{1} + t_{2} + ldots + t_{n} rightarrow text{max}$$
$$t_{1}^{2} + t_{2}^{2} + ldots + t_{n}^{2} = 1$$
$$t_{i} geq 0, forall i = 1, ldots, n$$
The Lagrangian for the problem is:
$$L = (t_{1} + ldots + t_{n}) - lambda (t_{1}^{2} + ldots + t_{n}^{2} - 1)$$
and $lambda$ stays for the lagrange multiplier.
The necessary extremum condition implies:
$$frac{partial L}{partial t_{i}} := 1 - lambda(2t_{i}) = 0$$
thus $t_{i} = frac{1}{2 lambda}$. Since $t_{i} geq 0$ it follows that $lambda > 0$.
The stationary point is $x = (frac{1}{2 lambda}, ldots, frac{1}{2 lambda})$ and we shall find the lambda such that the solution satisfy the second condition in the problem above.
$||x||_{2} = 1$ is equivalent to
$$ n cdot frac{1}{4 lambda^{2}} = 1$$ from which we get
$lambda = frac{sqrt{n}}{2}$
(note that we shall choose it positive due to the restrictions mentioned above)
Thus $t_{i} = frac{1}{sqrt{n}}$ and the maximum equals $$text{max} = n cdot frac{1}{sqrt{n}} = sqrt{n}$$
answered Dec 16 '18 at 20:52
hyperkahlerhyperkahler
1,483714
$endgroup$
I will elaborate on my comment above
Given an operator $T: X rightarrow Y$ where $$X = mathbb{K}^{n}$$
$$Y = l^{1}(mathbb{N})$$ its norm is given by
$$||T||_{op} = sup_{x neq 0}{frac{||Tx||_{1}}{||x||_{2}}} = sup_{||x||_{2} leq 1}{||Tx||_{1}} = sup_{||x||_{2} = 1}{||Tx||_{1}}$$
So we have to maximize $$||Tx||_{1} = |x_{1}| + ldots + |x_{n}|$$ given $$||x||_{2} := (x_{1}^{2} + ldots + x_{n}^{2})^{frac{1}{2}} = 1$$
Let $t_{i} := |x_{i}|$, then our problem reformulates as follows:
$$t_{1} + t_{2} + ldots + t_{n} rightarrow text{max}$$
$$t_{1}^{2} + t_{2}^{2} + ldots + t_{n}^{2} = 1$$
$$t_{i} geq 0, forall i = 1, ldots, n$$
The Lagrangian for the problem is:
$$L = (t_{1} + ldots + t_{n}) - lambda (t_{1}^{2} + ldots + t_{n}^{2} - 1)$$
and $lambda$ stays for the lagrange multiplier.
The necessary extremum condition implies:
$$frac{partial L}{partial t_{i}} := 1 - lambda(2t_{i}) = 0$$
thus $t_{i} = frac{1}{2 lambda}$. Since $t_{i} geq 0$ it follows that $lambda > 0$.
The stationary point is $x = (frac{1}{2 lambda}, ldots, frac{1}{2 lambda})$ and we shall find the lambda such that the solution satisfy the second condition in the problem above.
$||x||_{2} = 1$ is equivalent to
$$ n cdot frac{1}{4 lambda^{2}} = 1$$ from which we get
$lambda = frac{sqrt{n}}{2}$
(note that we shall choose it positive due to the restrictions mentioned above)
Thus $t_{i} = frac{1}{sqrt{n}}$ and the maximum equals $$text{max} = n cdot frac{1}{sqrt{n}} = sqrt{n}$$
answered Dec 16 '18 at 20:52
hyperkahlerhyperkahler
1,483714
edited Dec 16 '18 at 21:24
answered Dec 16 '18 at 20:52
hyperkahlerhyperkahler
1,483714
answered Dec 16 '18 at 20:52
hyperkahlerhyperkahler
1,483714
answered Dec 16 '18 at 20:52
hyperkahlerhyperkahler
1,483714
1,483714
1
$begingroup$
Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
$endgroup$
– Rebellos
Dec 16 '18 at 21:16
$begingroup$
@Rebellos i've made a mistake, please, check the answer for the updates
$endgroup$
– hyperkahler
Dec 16 '18 at 21:25
add a comment |
1
$begingroup$
Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
$endgroup$
– Rebellos
Dec 16 '18 at 21:16
$begingroup$
@Rebellos i've made a mistake, please, check the answer for the updates
$endgroup$
– hyperkahler
Dec 16 '18 at 21:25
1
1
$begingroup$
Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
$endgroup$
– Rebellos
Dec 16 '18 at 21:16
$begingroup$
Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
$endgroup$
– Rebellos
Dec 16 '18 at 21:16
$begingroup$
@Rebellos i've made a mistake, please, check the answer for the updates
$endgroup$
– hyperkahler
Dec 16 '18 at 21:25
$begingroup$
@Rebellos i've made a mistake, please, check the answer for the updates
$endgroup$
– hyperkahler
Dec 16 '18 at 21:25
add a comment |
$begingroup$
Using Cauchy-Schwarz inequality, we have
$$
sum_{i=1}^nleftlvert x_irightrvert=sum_{i=1}^nleftlvert x_irightrvertcdot 1leqslant left(sum_{i=1}^nx_i^2right)^{1/2}left(sum_{i=1}^n1right)^{1/2}=sqrt nleft(sum_{i=1}^nx_i^2right)^{1/2}
$$
hence $leftlVert TrightrVertleqslant sqrt n$.
For the opposite inequality, look at the case where $x_i=1$ for all $iin{1,dots,n}$.
answered Dec 18 '18 at 10:56
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
add a comment |
$begingroup$
Using Cauchy-Schwarz inequality, we have
$$
sum_{i=1}^nleftlvert x_irightrvert=sum_{i=1}^nleftlvert x_irightrvertcdot 1leqslant left(sum_{i=1}^nx_i^2right)^{1/2}left(sum_{i=1}^n1right)^{1/2}=sqrt nleft(sum_{i=1}^nx_i^2right)^{1/2}
$$
hence $leftlVert TrightrVertleqslant sqrt n$.
For the opposite inequality, look at the case where $x_i=1$ for all $iin{1,dots,n}$.
answered Dec 18 '18 at 10:56
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
add a comment |
$begingroup$
Using Cauchy-Schwarz inequality, we have
$$
sum_{i=1}^nleftlvert x_irightrvert=sum_{i=1}^nleftlvert x_irightrvertcdot 1leqslant left(sum_{i=1}^nx_i^2right)^{1/2}left(sum_{i=1}^n1right)^{1/2}=sqrt nleft(sum_{i=1}^nx_i^2right)^{1/2}
$$
hence $leftlVert TrightrVertleqslant sqrt n$.
For the opposite inequality, look at the case where $x_i=1$ for all $iin{1,dots,n}$.
answered Dec 18 '18 at 10:56
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
Using Cauchy-Schwarz inequality, we have
$$
sum_{i=1}^nleftlvert x_irightrvert=sum_{i=1}^nleftlvert x_irightrvertcdot 1leqslant left(sum_{i=1}^nx_i^2right)^{1/2}left(sum_{i=1}^n1right)^{1/2}=sqrt nleft(sum_{i=1}^nx_i^2right)^{1/2}
$$
hence $leftlVert TrightrVertleqslant sqrt n$.
For the opposite inequality, look at the case where $x_i=1$ for all $iin{1,dots,n}$.
answered Dec 18 '18 at 10:56
Davide GiraudoDavide Giraudo
127k16151265
edited Dec 18 '18 at 15:41
answered Dec 18 '18 at 10:56
Davide GiraudoDavide Giraudo
127k16151265
answered Dec 18 '18 at 10:56
Davide GiraudoDavide Giraudo
127k16151265
answered Dec 18 '18 at 10:56
Davide GiraudoDavide Giraudo
127k16151265
127k16151265
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$begingroup$
I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
$endgroup$
– hyperkahler
Dec 16 '18 at 20:30
$begingroup$
I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
$endgroup$
– Anna Schmitz
Dec 16 '18 at 20:38