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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The Binding of Isaac: Rebirth/Afterbirth

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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me
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