Is the norm $p(x) = max_{t in [0,2]}|x(t)| + left(int_0^1|x(t)|^7right)^{1/7}$ on $C[0,2]$ induced by any...












0












$begingroup$



I have a norm on space $X = C[0,2]$:
$$p(x) = max_{t in [0,2]}|x(t)| + left(int_0^1|x(t)|^7right)^{1/7}$$
Is that norm induced by any scalar product?




I try to find counterexample for parallelogram identity
$$||f+g||^2 + ||f-g||^2 = 2(||f||^2 + ||g||^2)$$
Why I tried to find it? Because I know the norm
$$d(x) = max_{t in [0,2]}|x(t)|$$
can't be induced by any scalar product. Maybe I should use polar identity: $$(f,g) = frac{1}{4}(||f+g||^2 + ||f-g||^2)$$



And then what to do with the norm, where instead of the integral - the sum? For example on space $C[-1,1]$:
$$k(x) = max_{t in [-1,1]}pi^{2t}|x(t)| + 5left(sum_{k=1}^{infty}frac{left|x(-1+frac{3}{2k})right|^2}{k^5}right)^{1/2}$$



Here, too, nothing happened. I know that a sum can be obtained from the scalar product (if $p=2$), if we consider it as a separate norm. And the maximum is impossible. How to work with such norms, where one of two norm can be induced?



Can you tell me a trick or something? Thank!



P.S (Edit): Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?






functional-analysis norm normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Trying to find a counterexample for parallelogram identity is a good idea. And that is not working?
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 14:32










  • $begingroup$
    I considered such functions that the integral was well considered. I tried about five examples that came to mind, but none of them turned out to be a counterexample.
    $endgroup$
    – mathmaniac
    Jan 19 at 14:34










  • $begingroup$
    It seems to be a counter example, used for max separately: $f = cos(frac{{pi}t}{4})$, $g=sin(frac{{pi}t}{4})$ But the integrals there are heavy, counted them with the help of wolframalpha and it seems like it happened. Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?
    $endgroup$
    – mathmaniac
    Jan 19 at 14:51








  • 1




    $begingroup$
    Surely some polynomials of low degree will be good counterexamples for the parallelogram law.
    $endgroup$
    – GEdgar
    Jan 19 at 15:00






  • 2




    $begingroup$
    for example $f(t)=1$ and $g(t)=t$.
    $endgroup$
    – GEdgar
    Jan 19 at 15:15
















0












$begingroup$



I have a norm on space $X = C[0,2]$:
$$p(x) = max_{t in [0,2]}|x(t)| + left(int_0^1|x(t)|^7right)^{1/7}$$
Is that norm induced by any scalar product?




I try to find counterexample for parallelogram identity
$$||f+g||^2 + ||f-g||^2 = 2(||f||^2 + ||g||^2)$$
Why I tried to find it? Because I know the norm
$$d(x) = max_{t in [0,2]}|x(t)|$$
can't be induced by any scalar product. Maybe I should use polar identity: $$(f,g) = frac{1}{4}(||f+g||^2 + ||f-g||^2)$$



And then what to do with the norm, where instead of the integral - the sum? For example on space $C[-1,1]$:
$$k(x) = max_{t in [-1,1]}pi^{2t}|x(t)| + 5left(sum_{k=1}^{infty}frac{left|x(-1+frac{3}{2k})right|^2}{k^5}right)^{1/2}$$



Here, too, nothing happened. I know that a sum can be obtained from the scalar product (if $p=2$), if we consider it as a separate norm. And the maximum is impossible. How to work with such norms, where one of two norm can be induced?



Can you tell me a trick or something? Thank!



P.S (Edit): Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?






functional-analysis norm normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Trying to find a counterexample for parallelogram identity is a good idea. And that is not working?
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 14:32










  • $begingroup$
    I considered such functions that the integral was well considered. I tried about five examples that came to mind, but none of them turned out to be a counterexample.
    $endgroup$
    – mathmaniac
    Jan 19 at 14:34










  • $begingroup$
    It seems to be a counter example, used for max separately: $f = cos(frac{{pi}t}{4})$, $g=sin(frac{{pi}t}{4})$ But the integrals there are heavy, counted them with the help of wolframalpha and it seems like it happened. Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?
    $endgroup$
    – mathmaniac
    Jan 19 at 14:51








  • 1




    $begingroup$
    Surely some polynomials of low degree will be good counterexamples for the parallelogram law.
    $endgroup$
    – GEdgar
    Jan 19 at 15:00






  • 2




    $begingroup$
    for example $f(t)=1$ and $g(t)=t$.
    $endgroup$
    – GEdgar
    Jan 19 at 15:15














0












0








0





$begingroup$



I have a norm on space $X = C[0,2]$:
$$p(x) = max_{t in [0,2]}|x(t)| + left(int_0^1|x(t)|^7right)^{1/7}$$
Is that norm induced by any scalar product?




I try to find counterexample for parallelogram identity
$$||f+g||^2 + ||f-g||^2 = 2(||f||^2 + ||g||^2)$$
Why I tried to find it? Because I know the norm
$$d(x) = max_{t in [0,2]}|x(t)|$$
can't be induced by any scalar product. Maybe I should use polar identity: $$(f,g) = frac{1}{4}(||f+g||^2 + ||f-g||^2)$$



And then what to do with the norm, where instead of the integral - the sum? For example on space $C[-1,1]$:
$$k(x) = max_{t in [-1,1]}pi^{2t}|x(t)| + 5left(sum_{k=1}^{infty}frac{left|x(-1+frac{3}{2k})right|^2}{k^5}right)^{1/2}$$



Here, too, nothing happened. I know that a sum can be obtained from the scalar product (if $p=2$), if we consider it as a separate norm. And the maximum is impossible. How to work with such norms, where one of two norm can be induced?



Can you tell me a trick or something? Thank!



P.S (Edit): Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?






functional-analysis norm normed-spaces







share|cite|improve this question











$endgroup$





I have a norm on space $X = C[0,2]$:
$$p(x) = max_{t in [0,2]}|x(t)| + left(int_0^1|x(t)|^7right)^{1/7}$$
Is that norm induced by any scalar product?




I try to find counterexample for parallelogram identity
$$||f+g||^2 + ||f-g||^2 = 2(||f||^2 + ||g||^2)$$
Why I tried to find it? Because I know the norm
$$d(x) = max_{t in [0,2]}|x(t)|$$
can't be induced by any scalar product. Maybe I should use polar identity: $$(f,g) = frac{1}{4}(||f+g||^2 + ||f-g||^2)$$



And then what to do with the norm, where instead of the integral - the sum? For example on space $C[-1,1]$:
$$k(x) = max_{t in [-1,1]}pi^{2t}|x(t)| + 5left(sum_{k=1}^{infty}frac{left|x(-1+frac{3}{2k})right|^2}{k^5}right)^{1/2}$$



Here, too, nothing happened. I know that a sum can be obtained from the scalar product (if $p=2$), if we consider it as a separate norm. And the maximum is impossible. How to work with such norms, where one of two norm can be induced?



Can you tell me a trick or something? Thank!



P.S (Edit): Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?






functional-analysis norm normed-spaces




functional-analysis norm normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 15:26









Blue

48.5k870154




48.5k870154










asked Jan 19 at 14:25









mathmaniacmathmaniac

18511




18511












  • $begingroup$
    Trying to find a counterexample for parallelogram identity is a good idea. And that is not working?
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 14:32










  • $begingroup$
    I considered such functions that the integral was well considered. I tried about five examples that came to mind, but none of them turned out to be a counterexample.
    $endgroup$
    – mathmaniac
    Jan 19 at 14:34










  • $begingroup$
    It seems to be a counter example, used for max separately: $f = cos(frac{{pi}t}{4})$, $g=sin(frac{{pi}t}{4})$ But the integrals there are heavy, counted them with the help of wolframalpha and it seems like it happened. Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?
    $endgroup$
    – mathmaniac
    Jan 19 at 14:51








  • 1




    $begingroup$
    Surely some polynomials of low degree will be good counterexamples for the parallelogram law.
    $endgroup$
    – GEdgar
    Jan 19 at 15:00






  • 2




    $begingroup$
    for example $f(t)=1$ and $g(t)=t$.
    $endgroup$
    – GEdgar
    Jan 19 at 15:15


















  • $begingroup$
    Trying to find a counterexample for parallelogram identity is a good idea. And that is not working?
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 14:32










  • $begingroup$
    I considered such functions that the integral was well considered. I tried about five examples that came to mind, but none of them turned out to be a counterexample.
    $endgroup$
    – mathmaniac
    Jan 19 at 14:34










  • $begingroup$
    It seems to be a counter example, used for max separately: $f = cos(frac{{pi}t}{4})$, $g=sin(frac{{pi}t}{4})$ But the integrals there are heavy, counted them with the help of wolframalpha and it seems like it happened. Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?
    $endgroup$
    – mathmaniac
    Jan 19 at 14:51








  • 1




    $begingroup$
    Surely some polynomials of low degree will be good counterexamples for the parallelogram law.
    $endgroup$
    – GEdgar
    Jan 19 at 15:00






  • 2




    $begingroup$
    for example $f(t)=1$ and $g(t)=t$.
    $endgroup$
    – GEdgar
    Jan 19 at 15:15
















$begingroup$
Trying to find a counterexample for parallelogram identity is a good idea. And that is not working?
$endgroup$
– mathcounterexamples.net
Jan 19 at 14:32




$begingroup$
Trying to find a counterexample for parallelogram identity is a good idea. And that is not working?
$endgroup$
– mathcounterexamples.net
Jan 19 at 14:32












$begingroup$
I considered such functions that the integral was well considered. I tried about five examples that came to mind, but none of them turned out to be a counterexample.
$endgroup$
– mathmaniac
Jan 19 at 14:34




$begingroup$
I considered such functions that the integral was well considered. I tried about five examples that came to mind, but none of them turned out to be a counterexample.
$endgroup$
– mathmaniac
Jan 19 at 14:34












$begingroup$
It seems to be a counter example, used for max separately: $f = cos(frac{{pi}t}{4})$, $g=sin(frac{{pi}t}{4})$ But the integrals there are heavy, counted them with the help of wolframalpha and it seems like it happened. Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?
$endgroup$
– mathmaniac
Jan 19 at 14:51






$begingroup$
It seems to be a counter example, used for max separately: $f = cos(frac{{pi}t}{4})$, $g=sin(frac{{pi}t}{4})$ But the integrals there are heavy, counted them with the help of wolframalpha and it seems like it happened. Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?
$endgroup$
– mathmaniac
Jan 19 at 14:51






1




1




$begingroup$
Surely some polynomials of low degree will be good counterexamples for the parallelogram law.
$endgroup$
– GEdgar
Jan 19 at 15:00




$begingroup$
Surely some polynomials of low degree will be good counterexamples for the parallelogram law.
$endgroup$
– GEdgar
Jan 19 at 15:00




2




2




$begingroup$
for example $f(t)=1$ and $g(t)=t$.
$endgroup$
– GEdgar
Jan 19 at 15:15




$begingroup$
for example $f(t)=1$ and $g(t)=t$.
$endgroup$
– GEdgar
Jan 19 at 15:15










1 Answer
1






active

oldest

votes


















1












$begingroup$

Consider



$$f(x) = begin{cases}
cos pi x & 0 le x le 1/2\
0 & 1/2 le x le 2
end{cases}$$



and $g(x) = f(2-x)$. You have $f,g in mathcal C([0,2],mathbb R)$. Let
$$I = left(int_0^2|f(t)|^7right)^{frac{1}{7}}=left(int_0^2|g(t)|^7right)^{frac{1}{7}} = frac{1}{2}left(int_0^2|f(t)+g(t)|^7right)^{frac{1}{7}}=frac{1}{2}left(int_0^2|f(t)-g(t)|^7right)^{frac{1}{7}}$$ and notice
that
$$1=max_{t in [0,2]}vert f(t)vert=max_{t in [0,2]}vert g(t)vert=max_{t in [0,2]}vert f(t)+g(t)vert=max_{t in [0,2]}vert f(t)-g(t)vert$$



Hence



$$Vert f+g Vert^2+Vert f-g Vert^2 =2(1+2I)^2 neq 4(1+I)^2= left(Vert f Vert^2 + Vert g Vert^2right)^2$$



as $I < frac{1}{sqrt 2}$.



Which allows to conclude.



Using the parallelogram identity is a general way to decide if a norm is induced by an inner product according to THE JORDAN-VON NEUMANN THEOREM






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the elegant counterexample!
    $endgroup$
    – mathmaniac
    Jan 19 at 15:28











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Consider



$$f(x) = begin{cases}
cos pi x & 0 le x le 1/2\
0 & 1/2 le x le 2
end{cases}$$



and $g(x) = f(2-x)$. You have $f,g in mathcal C([0,2],mathbb R)$. Let
$$I = left(int_0^2|f(t)|^7right)^{frac{1}{7}}=left(int_0^2|g(t)|^7right)^{frac{1}{7}} = frac{1}{2}left(int_0^2|f(t)+g(t)|^7right)^{frac{1}{7}}=frac{1}{2}left(int_0^2|f(t)-g(t)|^7right)^{frac{1}{7}}$$ and notice
that
$$1=max_{t in [0,2]}vert f(t)vert=max_{t in [0,2]}vert g(t)vert=max_{t in [0,2]}vert f(t)+g(t)vert=max_{t in [0,2]}vert f(t)-g(t)vert$$



Hence



$$Vert f+g Vert^2+Vert f-g Vert^2 =2(1+2I)^2 neq 4(1+I)^2= left(Vert f Vert^2 + Vert g Vert^2right)^2$$



as $I < frac{1}{sqrt 2}$.



Which allows to conclude.



Using the parallelogram identity is a general way to decide if a norm is induced by an inner product according to THE JORDAN-VON NEUMANN THEOREM






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the elegant counterexample!
    $endgroup$
    – mathmaniac
    Jan 19 at 15:28
















1












$begingroup$

Consider



$$f(x) = begin{cases}
cos pi x & 0 le x le 1/2\
0 & 1/2 le x le 2
end{cases}$$



and $g(x) = f(2-x)$. You have $f,g in mathcal C([0,2],mathbb R)$. Let
$$I = left(int_0^2|f(t)|^7right)^{frac{1}{7}}=left(int_0^2|g(t)|^7right)^{frac{1}{7}} = frac{1}{2}left(int_0^2|f(t)+g(t)|^7right)^{frac{1}{7}}=frac{1}{2}left(int_0^2|f(t)-g(t)|^7right)^{frac{1}{7}}$$ and notice
that
$$1=max_{t in [0,2]}vert f(t)vert=max_{t in [0,2]}vert g(t)vert=max_{t in [0,2]}vert f(t)+g(t)vert=max_{t in [0,2]}vert f(t)-g(t)vert$$



Hence



$$Vert f+g Vert^2+Vert f-g Vert^2 =2(1+2I)^2 neq 4(1+I)^2= left(Vert f Vert^2 + Vert g Vert^2right)^2$$



as $I < frac{1}{sqrt 2}$.



Which allows to conclude.



Using the parallelogram identity is a general way to decide if a norm is induced by an inner product according to THE JORDAN-VON NEUMANN THEOREM






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the elegant counterexample!
    $endgroup$
    – mathmaniac
    Jan 19 at 15:28














1












1








1





$begingroup$

Consider



$$f(x) = begin{cases}
cos pi x & 0 le x le 1/2\
0 & 1/2 le x le 2
end{cases}$$



and $g(x) = f(2-x)$. You have $f,g in mathcal C([0,2],mathbb R)$. Let
$$I = left(int_0^2|f(t)|^7right)^{frac{1}{7}}=left(int_0^2|g(t)|^7right)^{frac{1}{7}} = frac{1}{2}left(int_0^2|f(t)+g(t)|^7right)^{frac{1}{7}}=frac{1}{2}left(int_0^2|f(t)-g(t)|^7right)^{frac{1}{7}}$$ and notice
that
$$1=max_{t in [0,2]}vert f(t)vert=max_{t in [0,2]}vert g(t)vert=max_{t in [0,2]}vert f(t)+g(t)vert=max_{t in [0,2]}vert f(t)-g(t)vert$$



Hence



$$Vert f+g Vert^2+Vert f-g Vert^2 =2(1+2I)^2 neq 4(1+I)^2= left(Vert f Vert^2 + Vert g Vert^2right)^2$$



as $I < frac{1}{sqrt 2}$.



Which allows to conclude.



Using the parallelogram identity is a general way to decide if a norm is induced by an inner product according to THE JORDAN-VON NEUMANN THEOREM






share|cite|improve this answer











$endgroup$



Consider



$$f(x) = begin{cases}
cos pi x & 0 le x le 1/2\
0 & 1/2 le x le 2
end{cases}$$



and $g(x) = f(2-x)$. You have $f,g in mathcal C([0,2],mathbb R)$. Let
$$I = left(int_0^2|f(t)|^7right)^{frac{1}{7}}=left(int_0^2|g(t)|^7right)^{frac{1}{7}} = frac{1}{2}left(int_0^2|f(t)+g(t)|^7right)^{frac{1}{7}}=frac{1}{2}left(int_0^2|f(t)-g(t)|^7right)^{frac{1}{7}}$$ and notice
that
$$1=max_{t in [0,2]}vert f(t)vert=max_{t in [0,2]}vert g(t)vert=max_{t in [0,2]}vert f(t)+g(t)vert=max_{t in [0,2]}vert f(t)-g(t)vert$$



Hence



$$Vert f+g Vert^2+Vert f-g Vert^2 =2(1+2I)^2 neq 4(1+I)^2= left(Vert f Vert^2 + Vert g Vert^2right)^2$$



as $I < frac{1}{sqrt 2}$.



Which allows to conclude.



Using the parallelogram identity is a general way to decide if a norm is induced by an inner product according to THE JORDAN-VON NEUMANN THEOREM







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 19 at 15:39

























answered Jan 19 at 15:23









mathcounterexamples.netmathcounterexamples.net

26.8k22157




26.8k22157












  • $begingroup$
    Thanks for the elegant counterexample!
    $endgroup$
    – mathmaniac
    Jan 19 at 15:28


















  • $begingroup$
    Thanks for the elegant counterexample!
    $endgroup$
    – mathmaniac
    Jan 19 at 15:28
















$begingroup$
Thanks for the elegant counterexample!
$endgroup$
– mathmaniac
Jan 19 at 15:28




$begingroup$
Thanks for the elegant counterexample!
$endgroup$
– mathmaniac
Jan 19 at 15:28


















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Antonio Litta Visconti Arese

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Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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