Calculating the norm of a specific bounded linear functional
$begingroup$
Let $L_1([0,1], m)$ be the Banach space of $mathbb{K}$-valued (i.e. $mathbb{C}$ or $mathbb{R}$-valued) Lebesgue-integrable functions, where $m$ is the Lebesgue measure, be equipped with the norm $‖f‖_1=int_{[0,1]}|f|,dm$ for $f in L_1([0,1], m)$. For $n geq 1$ define $g_n(x) =nsin(n^2x)$, for $x in [0,1]$.
Furthermore, let $phi_n(f)=int_{[0,1]} f g_n , dm$.
I'm then asked to show that $phi_n(f)$ defines a bounded linear map (which I've done) and to show $‖phi_n‖=n$ for $n geq 2$. What I've done is that I have shown $‖phi_n‖leq n$. If I could find a function $f in L_1([0,1], m)$ such that $|phi_n(f)|=n$ and $‖f‖_1=1$, then I would have shown equality. How can I find such a function?
The problem also asks whether I can find a function $f in L_1([0,1], m)$ such that $lim_{n rightarrow infty} |phi_n(f)|=infty$ (perhaps I could use the same function as above?)
functional-analysis lebesgue-measure lp-spaces
$endgroup$
add a comment |
$begingroup$
Let $L_1([0,1], m)$ be the Banach space of $mathbb{K}$-valued (i.e. $mathbb{C}$ or $mathbb{R}$-valued) Lebesgue-integrable functions, where $m$ is the Lebesgue measure, be equipped with the norm $‖f‖_1=int_{[0,1]}|f|,dm$ for $f in L_1([0,1], m)$. For $n geq 1$ define $g_n(x) =nsin(n^2x)$, for $x in [0,1]$.
Furthermore, let $phi_n(f)=int_{[0,1]} f g_n , dm$.
I'm then asked to show that $phi_n(f)$ defines a bounded linear map (which I've done) and to show $‖phi_n‖=n$ for $n geq 2$. What I've done is that I have shown $‖phi_n‖leq n$. If I could find a function $f in L_1([0,1], m)$ such that $|phi_n(f)|=n$ and $‖f‖_1=1$, then I would have shown equality. How can I find such a function?
The problem also asks whether I can find a function $f in L_1([0,1], m)$ such that $lim_{n rightarrow infty} |phi_n(f)|=infty$ (perhaps I could use the same function as above?)
functional-analysis lebesgue-measure lp-spaces
$endgroup$
add a comment |
$begingroup$
Let $L_1([0,1], m)$ be the Banach space of $mathbb{K}$-valued (i.e. $mathbb{C}$ or $mathbb{R}$-valued) Lebesgue-integrable functions, where $m$ is the Lebesgue measure, be equipped with the norm $‖f‖_1=int_{[0,1]}|f|,dm$ for $f in L_1([0,1], m)$. For $n geq 1$ define $g_n(x) =nsin(n^2x)$, for $x in [0,1]$.
Furthermore, let $phi_n(f)=int_{[0,1]} f g_n , dm$.
I'm then asked to show that $phi_n(f)$ defines a bounded linear map (which I've done) and to show $‖phi_n‖=n$ for $n geq 2$. What I've done is that I have shown $‖phi_n‖leq n$. If I could find a function $f in L_1([0,1], m)$ such that $|phi_n(f)|=n$ and $‖f‖_1=1$, then I would have shown equality. How can I find such a function?
The problem also asks whether I can find a function $f in L_1([0,1], m)$ such that $lim_{n rightarrow infty} |phi_n(f)|=infty$ (perhaps I could use the same function as above?)
functional-analysis lebesgue-measure lp-spaces
$endgroup$
Let $L_1([0,1], m)$ be the Banach space of $mathbb{K}$-valued (i.e. $mathbb{C}$ or $mathbb{R}$-valued) Lebesgue-integrable functions, where $m$ is the Lebesgue measure, be equipped with the norm $‖f‖_1=int_{[0,1]}|f|,dm$ for $f in L_1([0,1], m)$. For $n geq 1$ define $g_n(x) =nsin(n^2x)$, for $x in [0,1]$.
Furthermore, let $phi_n(f)=int_{[0,1]} f g_n , dm$.
I'm then asked to show that $phi_n(f)$ defines a bounded linear map (which I've done) and to show $‖phi_n‖=n$ for $n geq 2$. What I've done is that I have shown $‖phi_n‖leq n$. If I could find a function $f in L_1([0,1], m)$ such that $|phi_n(f)|=n$ and $‖f‖_1=1$, then I would have shown equality. How can I find such a function?
The problem also asks whether I can find a function $f in L_1([0,1], m)$ such that $lim_{n rightarrow infty} |phi_n(f)|=infty$ (perhaps I could use the same function as above?)
functional-analysis lebesgue-measure lp-spaces
functional-analysis lebesgue-measure lp-spaces
asked Dec 6 '18 at 22:01
VercingetorixVercingetorix
285
285
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add a comment |
1 Answer
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$begingroup$
It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $Lcolon mathbb L^1tomathbb R$ defined by $L(f)=int_{(0,1)}g(x)f(x)mathrm dlambda(x)$ is $leftlVert grightrVert_infty$. Indeed, fix a positive $varepsilon<leftlVert grightrVert_infty$ and define the sets $A^+:=left{xin (0,1), g(x)gt leftlVert grightrVert_infty-varepsilonright}$ and $A^-:=left{xin (0,1), -g(x)gt leftlVert grightrVert_infty-varepsilonright}$. Then
$$
maxleft{Lleft(frac 1{lambdaleft(A^+right)}mathbf 1_{A^+}right),Lleft(frac 1{lambdaleft(A^-right)}mathbf 1_{A^-}right) right}geqslant leftlVert grightrVert_infty-varepsilon
$$
and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.
For the other question, look at one of the functions $xmapsto pm x^{-alpha}$ for $1/2ltalphalt 1$.
$endgroup$
$begingroup$
I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:20
$begingroup$
but I don't know how to do it without explicitly finding a function
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:27
$begingroup$
The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
$endgroup$
– Davide Giraudo
Dec 11 '18 at 19:05
$begingroup$
Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
$endgroup$
– Vercingetorix
Dec 11 '18 at 19:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $Lcolon mathbb L^1tomathbb R$ defined by $L(f)=int_{(0,1)}g(x)f(x)mathrm dlambda(x)$ is $leftlVert grightrVert_infty$. Indeed, fix a positive $varepsilon<leftlVert grightrVert_infty$ and define the sets $A^+:=left{xin (0,1), g(x)gt leftlVert grightrVert_infty-varepsilonright}$ and $A^-:=left{xin (0,1), -g(x)gt leftlVert grightrVert_infty-varepsilonright}$. Then
$$
maxleft{Lleft(frac 1{lambdaleft(A^+right)}mathbf 1_{A^+}right),Lleft(frac 1{lambdaleft(A^-right)}mathbf 1_{A^-}right) right}geqslant leftlVert grightrVert_infty-varepsilon
$$
and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.
For the other question, look at one of the functions $xmapsto pm x^{-alpha}$ for $1/2ltalphalt 1$.
$endgroup$
$begingroup$
I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:20
$begingroup$
but I don't know how to do it without explicitly finding a function
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:27
$begingroup$
The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
$endgroup$
– Davide Giraudo
Dec 11 '18 at 19:05
$begingroup$
Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
$endgroup$
– Vercingetorix
Dec 11 '18 at 19:49
add a comment |
$begingroup$
It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $Lcolon mathbb L^1tomathbb R$ defined by $L(f)=int_{(0,1)}g(x)f(x)mathrm dlambda(x)$ is $leftlVert grightrVert_infty$. Indeed, fix a positive $varepsilon<leftlVert grightrVert_infty$ and define the sets $A^+:=left{xin (0,1), g(x)gt leftlVert grightrVert_infty-varepsilonright}$ and $A^-:=left{xin (0,1), -g(x)gt leftlVert grightrVert_infty-varepsilonright}$. Then
$$
maxleft{Lleft(frac 1{lambdaleft(A^+right)}mathbf 1_{A^+}right),Lleft(frac 1{lambdaleft(A^-right)}mathbf 1_{A^-}right) right}geqslant leftlVert grightrVert_infty-varepsilon
$$
and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.
For the other question, look at one of the functions $xmapsto pm x^{-alpha}$ for $1/2ltalphalt 1$.
$endgroup$
$begingroup$
I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:20
$begingroup$
but I don't know how to do it without explicitly finding a function
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:27
$begingroup$
The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
$endgroup$
– Davide Giraudo
Dec 11 '18 at 19:05
$begingroup$
Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
$endgroup$
– Vercingetorix
Dec 11 '18 at 19:49
add a comment |
$begingroup$
It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $Lcolon mathbb L^1tomathbb R$ defined by $L(f)=int_{(0,1)}g(x)f(x)mathrm dlambda(x)$ is $leftlVert grightrVert_infty$. Indeed, fix a positive $varepsilon<leftlVert grightrVert_infty$ and define the sets $A^+:=left{xin (0,1), g(x)gt leftlVert grightrVert_infty-varepsilonright}$ and $A^-:=left{xin (0,1), -g(x)gt leftlVert grightrVert_infty-varepsilonright}$. Then
$$
maxleft{Lleft(frac 1{lambdaleft(A^+right)}mathbf 1_{A^+}right),Lleft(frac 1{lambdaleft(A^-right)}mathbf 1_{A^-}right) right}geqslant leftlVert grightrVert_infty-varepsilon
$$
and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.
For the other question, look at one of the functions $xmapsto pm x^{-alpha}$ for $1/2ltalphalt 1$.
$endgroup$
It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $Lcolon mathbb L^1tomathbb R$ defined by $L(f)=int_{(0,1)}g(x)f(x)mathrm dlambda(x)$ is $leftlVert grightrVert_infty$. Indeed, fix a positive $varepsilon<leftlVert grightrVert_infty$ and define the sets $A^+:=left{xin (0,1), g(x)gt leftlVert grightrVert_infty-varepsilonright}$ and $A^-:=left{xin (0,1), -g(x)gt leftlVert grightrVert_infty-varepsilonright}$. Then
$$
maxleft{Lleft(frac 1{lambdaleft(A^+right)}mathbf 1_{A^+}right),Lleft(frac 1{lambdaleft(A^-right)}mathbf 1_{A^-}right) right}geqslant leftlVert grightrVert_infty-varepsilon
$$
and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.
For the other question, look at one of the functions $xmapsto pm x^{-alpha}$ for $1/2ltalphalt 1$.
answered Dec 10 '18 at 23:04
Davide GiraudoDavide Giraudo
127k17154268
127k17154268
$begingroup$
I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:20
$begingroup$
but I don't know how to do it without explicitly finding a function
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:27
$begingroup$
The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
$endgroup$
– Davide Giraudo
Dec 11 '18 at 19:05
$begingroup$
Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
$endgroup$
– Vercingetorix
Dec 11 '18 at 19:49
add a comment |
$begingroup$
I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:20
$begingroup$
but I don't know how to do it without explicitly finding a function
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:27
$begingroup$
The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
$endgroup$
– Davide Giraudo
Dec 11 '18 at 19:05
$begingroup$
Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
$endgroup$
– Vercingetorix
Dec 11 '18 at 19:49
$begingroup$
I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:20
$begingroup$
I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:20
$begingroup$
but I don't know how to do it without explicitly finding a function
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:27
$begingroup$
but I don't know how to do it without explicitly finding a function
$endgroup$
– Vercingetorix
Dec 11 '18 at 18:27
$begingroup$
The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
$endgroup$
– Davide Giraudo
Dec 11 '18 at 19:05
$begingroup$
The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
$endgroup$
– Davide Giraudo
Dec 11 '18 at 19:05
$begingroup$
Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
$endgroup$
– Vercingetorix
Dec 11 '18 at 19:49
$begingroup$
Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
$endgroup$
– Vercingetorix
Dec 11 '18 at 19:49
add a comment |
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