Prove that $max(||f||_{infty},||f'||_{infty})leq frac12 |f'|_{L^2([0,1])}$












2












$begingroup$


i want to prove this:
$$maxleft(||f||_{infty},||f'||_{infty}right)leq frac12 |f'|_{L^2([0,1])}$$
where $fin C^1([0,1])$ and $f(0)=f(1)=0$.



What I did is: (using Cauchy Schwartz inequality)
$$
|f(t)|=left|int_0^t f'(t) ,dtright|le int_0^t |f'(t)| ,dtleq left(int_0^1 |f'(t)|^2 ,dtright)^{frac12}=|f'|_{L^2([0,1])}$$



I don't know how to continue. Any help please?






analysis sobolev-spaces







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$endgroup$












  • $begingroup$
    Search "Poincare inequality" - it might help
    $endgroup$
    – Zachary Selk
    Dec 20 '18 at 20:14










  • $begingroup$
    @ZacharySelk I don't know how to find $frac12$
    $endgroup$
    – Vrouvrou
    Dec 20 '18 at 20:21
















2












$begingroup$


i want to prove this:
$$maxleft(||f||_{infty},||f'||_{infty}right)leq frac12 |f'|_{L^2([0,1])}$$
where $fin C^1([0,1])$ and $f(0)=f(1)=0$.



What I did is: (using Cauchy Schwartz inequality)
$$
|f(t)|=left|int_0^t f'(t) ,dtright|le int_0^t |f'(t)| ,dtleq left(int_0^1 |f'(t)|^2 ,dtright)^{frac12}=|f'|_{L^2([0,1])}$$



I don't know how to continue. Any help please?






analysis sobolev-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Search "Poincare inequality" - it might help
    $endgroup$
    – Zachary Selk
    Dec 20 '18 at 20:14










  • $begingroup$
    @ZacharySelk I don't know how to find $frac12$
    $endgroup$
    – Vrouvrou
    Dec 20 '18 at 20:21














2












2








2





$begingroup$


i want to prove this:
$$maxleft(||f||_{infty},||f'||_{infty}right)leq frac12 |f'|_{L^2([0,1])}$$
where $fin C^1([0,1])$ and $f(0)=f(1)=0$.



What I did is: (using Cauchy Schwartz inequality)
$$
|f(t)|=left|int_0^t f'(t) ,dtright|le int_0^t |f'(t)| ,dtleq left(int_0^1 |f'(t)|^2 ,dtright)^{frac12}=|f'|_{L^2([0,1])}$$



I don't know how to continue. Any help please?






analysis sobolev-spaces







share|cite|improve this question











$endgroup$




i want to prove this:
$$maxleft(||f||_{infty},||f'||_{infty}right)leq frac12 |f'|_{L^2([0,1])}$$
where $fin C^1([0,1])$ and $f(0)=f(1)=0$.



What I did is: (using Cauchy Schwartz inequality)
$$
|f(t)|=left|int_0^t f'(t) ,dtright|le int_0^t |f'(t)| ,dtleq left(int_0^1 |f'(t)|^2 ,dtright)^{frac12}=|f'|_{L^2([0,1])}$$



I don't know how to continue. Any help please?






analysis sobolev-spaces




analysis sobolev-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 9:25







Vrouvrou

















asked Dec 20 '18 at 20:08









VrouvrouVrouvrou

1,9191722




1,9191722












  • $begingroup$
    Search "Poincare inequality" - it might help
    $endgroup$
    – Zachary Selk
    Dec 20 '18 at 20:14










  • $begingroup$
    @ZacharySelk I don't know how to find $frac12$
    $endgroup$
    – Vrouvrou
    Dec 20 '18 at 20:21


















  • $begingroup$
    Search "Poincare inequality" - it might help
    $endgroup$
    – Zachary Selk
    Dec 20 '18 at 20:14










  • $begingroup$
    @ZacharySelk I don't know how to find $frac12$
    $endgroup$
    – Vrouvrou
    Dec 20 '18 at 20:21
















$begingroup$
Search "Poincare inequality" - it might help
$endgroup$
– Zachary Selk
Dec 20 '18 at 20:14




$begingroup$
Search "Poincare inequality" - it might help
$endgroup$
– Zachary Selk
Dec 20 '18 at 20:14












$begingroup$
@ZacharySelk I don't know how to find $frac12$
$endgroup$
– Vrouvrou
Dec 20 '18 at 20:21




$begingroup$
@ZacharySelk I don't know how to find $frac12$
$endgroup$
– Vrouvrou
Dec 20 '18 at 20:21










2 Answers
2






active

oldest

votes


















1












$begingroup$

I don't think that the inequality holds. For example, let $f(t)=sin(2pi t)$. Then
$$ fin C^1[0,1], f(0)=f(1)=0. $$
Clearly
$$ max_{tin[0,1]}|f'(t)|=2pi,|f'|_{L^2[0,1]}=sqrt2pi. $$
But
$$ max_{tin[0,1]}|f'(t)|>frac12|f'|_{L^2[0,1]}. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    For extra credit you can try to show $max |f'| le M |f'|_2$ fails in general for any positive $M$.
    $endgroup$
    – Umberto P.
    Dec 20 '18 at 21:13










  • $begingroup$
    @UmbertoP., you're right.
    $endgroup$
    – xpaul
    Dec 20 '18 at 21:15



















1












$begingroup$

You were on the right track, but you didn't use $f(1) = 0$.



For $t in [0,1]$ we have
begin{align}
|f(t)| &= frac12 left|int_0^t f'(t),dtright| + frac12left|int_t^1 f'(t),dtright| \
&leq frac12 int_0^t |f'(t)|,dt + frac12int_t^1 |f'(t)|,dt\
&= frac12 int_0^1 |f'(t)|,dt \
&le
frac12left(int_0^1 |f'(t)|^2 ,dtright)^{frac12}\
&=frac12|f'|_{L^2([0,1])}
end{align}
again using Cauchy-Schwartz. Therefore $|f|_infty le frac12|f'|_{L^2([0,1])}$.



The inequality $|f'|_infty le frac12|f'|_{L^2([0,1])}$ isn't true.



As suggested by @Martin R, consider $f(x) = x(1-x)$. We have $|f'|_infty = 1$ but $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Unless I am mistaken, your first part gives an estimate for $|f|_infty$, not for $|f'|_infty $. That is a part of the intended inequality, but $|f'|_infty le frac12|f'|_{L^2([0,1])}$ remains to show. Or am I overlooking something?
    $endgroup$
    – Martin R
    Dec 20 '18 at 20:36












  • $begingroup$
    @MartinR You're right, corrected. I'm not yet sure how to prove the other inequality.
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 20:38








  • 1




    $begingroup$
    Actually the estimate for $f'$ seems to be wrong, $f(x) = x(1-x)$ looks like a counter-example to me.
    $endgroup$
    – Martin R
    Dec 20 '18 at 21:06












  • $begingroup$
    @MartinR in fact it can't hold with any constant other than $frac 12$ either.
    $endgroup$
    – Umberto P.
    Dec 20 '18 at 21:12










  • $begingroup$
    @MartinR Are you sure? We have $|f|_infty = frac14$ and $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}} = 0.288675...$
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 21:13











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

I don't think that the inequality holds. For example, let $f(t)=sin(2pi t)$. Then
$$ fin C^1[0,1], f(0)=f(1)=0. $$
Clearly
$$ max_{tin[0,1]}|f'(t)|=2pi,|f'|_{L^2[0,1]}=sqrt2pi. $$
But
$$ max_{tin[0,1]}|f'(t)|>frac12|f'|_{L^2[0,1]}. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    For extra credit you can try to show $max |f'| le M |f'|_2$ fails in general for any positive $M$.
    $endgroup$
    – Umberto P.
    Dec 20 '18 at 21:13










  • $begingroup$
    @UmbertoP., you're right.
    $endgroup$
    – xpaul
    Dec 20 '18 at 21:15
















1












$begingroup$

I don't think that the inequality holds. For example, let $f(t)=sin(2pi t)$. Then
$$ fin C^1[0,1], f(0)=f(1)=0. $$
Clearly
$$ max_{tin[0,1]}|f'(t)|=2pi,|f'|_{L^2[0,1]}=sqrt2pi. $$
But
$$ max_{tin[0,1]}|f'(t)|>frac12|f'|_{L^2[0,1]}. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    For extra credit you can try to show $max |f'| le M |f'|_2$ fails in general for any positive $M$.
    $endgroup$
    – Umberto P.
    Dec 20 '18 at 21:13










  • $begingroup$
    @UmbertoP., you're right.
    $endgroup$
    – xpaul
    Dec 20 '18 at 21:15














1












1








1





$begingroup$

I don't think that the inequality holds. For example, let $f(t)=sin(2pi t)$. Then
$$ fin C^1[0,1], f(0)=f(1)=0. $$
Clearly
$$ max_{tin[0,1]}|f'(t)|=2pi,|f'|_{L^2[0,1]}=sqrt2pi. $$
But
$$ max_{tin[0,1]}|f'(t)|>frac12|f'|_{L^2[0,1]}. $$






share|cite|improve this answer









$endgroup$



I don't think that the inequality holds. For example, let $f(t)=sin(2pi t)$. Then
$$ fin C^1[0,1], f(0)=f(1)=0. $$
Clearly
$$ max_{tin[0,1]}|f'(t)|=2pi,|f'|_{L^2[0,1]}=sqrt2pi. $$
But
$$ max_{tin[0,1]}|f'(t)|>frac12|f'|_{L^2[0,1]}. $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 21:12









xpaulxpaul

22.6k24455




22.6k24455












  • $begingroup$
    For extra credit you can try to show $max |f'| le M |f'|_2$ fails in general for any positive $M$.
    $endgroup$
    – Umberto P.
    Dec 20 '18 at 21:13










  • $begingroup$
    @UmbertoP., you're right.
    $endgroup$
    – xpaul
    Dec 20 '18 at 21:15


















  • $begingroup$
    For extra credit you can try to show $max |f'| le M |f'|_2$ fails in general for any positive $M$.
    $endgroup$
    – Umberto P.
    Dec 20 '18 at 21:13










  • $begingroup$
    @UmbertoP., you're right.
    $endgroup$
    – xpaul
    Dec 20 '18 at 21:15
















$begingroup$
For extra credit you can try to show $max |f'| le M |f'|_2$ fails in general for any positive $M$.
$endgroup$
– Umberto P.
Dec 20 '18 at 21:13




$begingroup$
For extra credit you can try to show $max |f'| le M |f'|_2$ fails in general for any positive $M$.
$endgroup$
– Umberto P.
Dec 20 '18 at 21:13












$begingroup$
@UmbertoP., you're right.
$endgroup$
– xpaul
Dec 20 '18 at 21:15




$begingroup$
@UmbertoP., you're right.
$endgroup$
– xpaul
Dec 20 '18 at 21:15











1












$begingroup$

You were on the right track, but you didn't use $f(1) = 0$.



For $t in [0,1]$ we have
begin{align}
|f(t)| &= frac12 left|int_0^t f'(t),dtright| + frac12left|int_t^1 f'(t),dtright| \
&leq frac12 int_0^t |f'(t)|,dt + frac12int_t^1 |f'(t)|,dt\
&= frac12 int_0^1 |f'(t)|,dt \
&le
frac12left(int_0^1 |f'(t)|^2 ,dtright)^{frac12}\
&=frac12|f'|_{L^2([0,1])}
end{align}
again using Cauchy-Schwartz. Therefore $|f|_infty le frac12|f'|_{L^2([0,1])}$.



The inequality $|f'|_infty le frac12|f'|_{L^2([0,1])}$ isn't true.



As suggested by @Martin R, consider $f(x) = x(1-x)$. We have $|f'|_infty = 1$ but $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Unless I am mistaken, your first part gives an estimate for $|f|_infty$, not for $|f'|_infty $. That is a part of the intended inequality, but $|f'|_infty le frac12|f'|_{L^2([0,1])}$ remains to show. Or am I overlooking something?
    $endgroup$
    – Martin R
    Dec 20 '18 at 20:36












  • $begingroup$
    @MartinR You're right, corrected. I'm not yet sure how to prove the other inequality.
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 20:38








  • 1




    $begingroup$
    Actually the estimate for $f'$ seems to be wrong, $f(x) = x(1-x)$ looks like a counter-example to me.
    $endgroup$
    – Martin R
    Dec 20 '18 at 21:06












  • $begingroup$
    @MartinR in fact it can't hold with any constant other than $frac 12$ either.
    $endgroup$
    – Umberto P.
    Dec 20 '18 at 21:12










  • $begingroup$
    @MartinR Are you sure? We have $|f|_infty = frac14$ and $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}} = 0.288675...$
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 21:13
















1












$begingroup$

You were on the right track, but you didn't use $f(1) = 0$.



For $t in [0,1]$ we have
begin{align}
|f(t)| &= frac12 left|int_0^t f'(t),dtright| + frac12left|int_t^1 f'(t),dtright| \
&leq frac12 int_0^t |f'(t)|,dt + frac12int_t^1 |f'(t)|,dt\
&= frac12 int_0^1 |f'(t)|,dt \
&le
frac12left(int_0^1 |f'(t)|^2 ,dtright)^{frac12}\
&=frac12|f'|_{L^2([0,1])}
end{align}
again using Cauchy-Schwartz. Therefore $|f|_infty le frac12|f'|_{L^2([0,1])}$.



The inequality $|f'|_infty le frac12|f'|_{L^2([0,1])}$ isn't true.



As suggested by @Martin R, consider $f(x) = x(1-x)$. We have $|f'|_infty = 1$ but $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Unless I am mistaken, your first part gives an estimate for $|f|_infty$, not for $|f'|_infty $. That is a part of the intended inequality, but $|f'|_infty le frac12|f'|_{L^2([0,1])}$ remains to show. Or am I overlooking something?
    $endgroup$
    – Martin R
    Dec 20 '18 at 20:36












  • $begingroup$
    @MartinR You're right, corrected. I'm not yet sure how to prove the other inequality.
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 20:38








  • 1




    $begingroup$
    Actually the estimate for $f'$ seems to be wrong, $f(x) = x(1-x)$ looks like a counter-example to me.
    $endgroup$
    – Martin R
    Dec 20 '18 at 21:06












  • $begingroup$
    @MartinR in fact it can't hold with any constant other than $frac 12$ either.
    $endgroup$
    – Umberto P.
    Dec 20 '18 at 21:12










  • $begingroup$
    @MartinR Are you sure? We have $|f|_infty = frac14$ and $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}} = 0.288675...$
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 21:13














1












1








1





$begingroup$

You were on the right track, but you didn't use $f(1) = 0$.



For $t in [0,1]$ we have
begin{align}
|f(t)| &= frac12 left|int_0^t f'(t),dtright| + frac12left|int_t^1 f'(t),dtright| \
&leq frac12 int_0^t |f'(t)|,dt + frac12int_t^1 |f'(t)|,dt\
&= frac12 int_0^1 |f'(t)|,dt \
&le
frac12left(int_0^1 |f'(t)|^2 ,dtright)^{frac12}\
&=frac12|f'|_{L^2([0,1])}
end{align}
again using Cauchy-Schwartz. Therefore $|f|_infty le frac12|f'|_{L^2([0,1])}$.



The inequality $|f'|_infty le frac12|f'|_{L^2([0,1])}$ isn't true.



As suggested by @Martin R, consider $f(x) = x(1-x)$. We have $|f'|_infty = 1$ but $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}}$.






share|cite|improve this answer











$endgroup$



You were on the right track, but you didn't use $f(1) = 0$.



For $t in [0,1]$ we have
begin{align}
|f(t)| &= frac12 left|int_0^t f'(t),dtright| + frac12left|int_t^1 f'(t),dtright| \
&leq frac12 int_0^t |f'(t)|,dt + frac12int_t^1 |f'(t)|,dt\
&= frac12 int_0^1 |f'(t)|,dt \
&le
frac12left(int_0^1 |f'(t)|^2 ,dtright)^{frac12}\
&=frac12|f'|_{L^2([0,1])}
end{align}
again using Cauchy-Schwartz. Therefore $|f|_infty le frac12|f'|_{L^2([0,1])}$.



The inequality $|f'|_infty le frac12|f'|_{L^2([0,1])}$ isn't true.



As suggested by @Martin R, consider $f(x) = x(1-x)$. We have $|f'|_infty = 1$ but $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 21:17

























answered Dec 20 '18 at 20:32









mechanodroidmechanodroid

27.3k62447




27.3k62447












  • $begingroup$
    Unless I am mistaken, your first part gives an estimate for $|f|_infty$, not for $|f'|_infty $. That is a part of the intended inequality, but $|f'|_infty le frac12|f'|_{L^2([0,1])}$ remains to show. Or am I overlooking something?
    $endgroup$
    – Martin R
    Dec 20 '18 at 20:36












  • $begingroup$
    @MartinR You're right, corrected. I'm not yet sure how to prove the other inequality.
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 20:38








  • 1




    $begingroup$
    Actually the estimate for $f'$ seems to be wrong, $f(x) = x(1-x)$ looks like a counter-example to me.
    $endgroup$
    – Martin R
    Dec 20 '18 at 21:06












  • $begingroup$
    @MartinR in fact it can't hold with any constant other than $frac 12$ either.
    $endgroup$
    – Umberto P.
    Dec 20 '18 at 21:12










  • $begingroup$
    @MartinR Are you sure? We have $|f|_infty = frac14$ and $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}} = 0.288675...$
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 21:13


















  • $begingroup$
    Unless I am mistaken, your first part gives an estimate for $|f|_infty$, not for $|f'|_infty $. That is a part of the intended inequality, but $|f'|_infty le frac12|f'|_{L^2([0,1])}$ remains to show. Or am I overlooking something?
    $endgroup$
    – Martin R
    Dec 20 '18 at 20:36












  • $begingroup$
    @MartinR You're right, corrected. I'm not yet sure how to prove the other inequality.
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 20:38








  • 1




    $begingroup$
    Actually the estimate for $f'$ seems to be wrong, $f(x) = x(1-x)$ looks like a counter-example to me.
    $endgroup$
    – Martin R
    Dec 20 '18 at 21:06












  • $begingroup$
    @MartinR in fact it can't hold with any constant other than $frac 12$ either.
    $endgroup$
    – Umberto P.
    Dec 20 '18 at 21:12










  • $begingroup$
    @MartinR Are you sure? We have $|f|_infty = frac14$ and $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}} = 0.288675...$
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 21:13
















$begingroup$
Unless I am mistaken, your first part gives an estimate for $|f|_infty$, not for $|f'|_infty $. That is a part of the intended inequality, but $|f'|_infty le frac12|f'|_{L^2([0,1])}$ remains to show. Or am I overlooking something?
$endgroup$
– Martin R
Dec 20 '18 at 20:36






$begingroup$
Unless I am mistaken, your first part gives an estimate for $|f|_infty$, not for $|f'|_infty $. That is a part of the intended inequality, but $|f'|_infty le frac12|f'|_{L^2([0,1])}$ remains to show. Or am I overlooking something?
$endgroup$
– Martin R
Dec 20 '18 at 20:36














$begingroup$
@MartinR You're right, corrected. I'm not yet sure how to prove the other inequality.
$endgroup$
– mechanodroid
Dec 20 '18 at 20:38






$begingroup$
@MartinR You're right, corrected. I'm not yet sure how to prove the other inequality.
$endgroup$
– mechanodroid
Dec 20 '18 at 20:38






1




1




$begingroup$
Actually the estimate for $f'$ seems to be wrong, $f(x) = x(1-x)$ looks like a counter-example to me.
$endgroup$
– Martin R
Dec 20 '18 at 21:06






$begingroup$
Actually the estimate for $f'$ seems to be wrong, $f(x) = x(1-x)$ looks like a counter-example to me.
$endgroup$
– Martin R
Dec 20 '18 at 21:06














$begingroup$
@MartinR in fact it can't hold with any constant other than $frac 12$ either.
$endgroup$
– Umberto P.
Dec 20 '18 at 21:12




$begingroup$
@MartinR in fact it can't hold with any constant other than $frac 12$ either.
$endgroup$
– Umberto P.
Dec 20 '18 at 21:12












$begingroup$
@MartinR Are you sure? We have $|f|_infty = frac14$ and $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}} = 0.288675...$
$endgroup$
– mechanodroid
Dec 20 '18 at 21:13




$begingroup$
@MartinR Are you sure? We have $|f|_infty = frac14$ and $frac12|f'|_{L^2([0,1])} = frac1{2sqrt{3}} = 0.288675...$
$endgroup$
– mechanodroid
Dec 20 '18 at 21:13


















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