Show that the functional is continuous












-1












$begingroup$


Task:



Check if the functional F: $$F(f) = int_{-1}^1 f(t) sgn(t) dt.$$is continuous on the space $mathbb E=L_2(-1, 1) $:



Solution:



I need to find $M>0$ and show the inequality: $$vert F(f) vert le M (int_{-1}^1 vert f(t)vert^2 dt)^{1/2}.$$



I started from $$ vert F(f) vert = vert int_{-1}^1 f(t) sgn(t) dt vert$$



But actually I do not know how to evaluate this function.



I know that somewhere I need to use the Hölder's inequality: $$ int_{-1}^1 vert f(t) g(t) vert dt lt (vert f(t) vert^2 dt)^{1/2} (vert g(t) vert^2 dt)^{1/2},$$






functional-analysis continuity holder-inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    What would be $g(t)$ in the Holder inequality?
    $endgroup$
    – Federico
    Dec 11 '18 at 18:01










  • $begingroup$
    I need to find $g(t)$ evaluating the function I started from
    $endgroup$
    – Begri
    Dec 11 '18 at 18:07










  • $begingroup$
    Yes, but for $f(t)mathrm{sign}(t)$ what $g$ do you take? It's quite obviuous... (And it's already spoiled in the answer...)
    $endgroup$
    – Federico
    Dec 11 '18 at 18:08


















-1












$begingroup$


Task:



Check if the functional F: $$F(f) = int_{-1}^1 f(t) sgn(t) dt.$$is continuous on the space $mathbb E=L_2(-1, 1) $:



Solution:



I need to find $M>0$ and show the inequality: $$vert F(f) vert le M (int_{-1}^1 vert f(t)vert^2 dt)^{1/2}.$$



I started from $$ vert F(f) vert = vert int_{-1}^1 f(t) sgn(t) dt vert$$



But actually I do not know how to evaluate this function.



I know that somewhere I need to use the Hölder's inequality: $$ int_{-1}^1 vert f(t) g(t) vert dt lt (vert f(t) vert^2 dt)^{1/2} (vert g(t) vert^2 dt)^{1/2},$$






functional-analysis continuity holder-inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    What would be $g(t)$ in the Holder inequality?
    $endgroup$
    – Federico
    Dec 11 '18 at 18:01










  • $begingroup$
    I need to find $g(t)$ evaluating the function I started from
    $endgroup$
    – Begri
    Dec 11 '18 at 18:07










  • $begingroup$
    Yes, but for $f(t)mathrm{sign}(t)$ what $g$ do you take? It's quite obviuous... (And it's already spoiled in the answer...)
    $endgroup$
    – Federico
    Dec 11 '18 at 18:08
















-1












-1








-1





$begingroup$


Task:



Check if the functional F: $$F(f) = int_{-1}^1 f(t) sgn(t) dt.$$is continuous on the space $mathbb E=L_2(-1, 1) $:



Solution:



I need to find $M>0$ and show the inequality: $$vert F(f) vert le M (int_{-1}^1 vert f(t)vert^2 dt)^{1/2}.$$



I started from $$ vert F(f) vert = vert int_{-1}^1 f(t) sgn(t) dt vert$$



But actually I do not know how to evaluate this function.



I know that somewhere I need to use the Hölder's inequality: $$ int_{-1}^1 vert f(t) g(t) vert dt lt (vert f(t) vert^2 dt)^{1/2} (vert g(t) vert^2 dt)^{1/2},$$






functional-analysis continuity holder-inequality







share|cite|improve this question









$endgroup$




Task:



Check if the functional F: $$F(f) = int_{-1}^1 f(t) sgn(t) dt.$$is continuous on the space $mathbb E=L_2(-1, 1) $:



Solution:



I need to find $M>0$ and show the inequality: $$vert F(f) vert le M (int_{-1}^1 vert f(t)vert^2 dt)^{1/2}.$$



I started from $$ vert F(f) vert = vert int_{-1}^1 f(t) sgn(t) dt vert$$



But actually I do not know how to evaluate this function.



I know that somewhere I need to use the Hölder's inequality: $$ int_{-1}^1 vert f(t) g(t) vert dt lt (vert f(t) vert^2 dt)^{1/2} (vert g(t) vert^2 dt)^{1/2},$$






functional-analysis continuity holder-inequality




functional-analysis continuity holder-inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 17:57









BegriBegri

83




83












  • $begingroup$
    What would be $g(t)$ in the Holder inequality?
    $endgroup$
    – Federico
    Dec 11 '18 at 18:01










  • $begingroup$
    I need to find $g(t)$ evaluating the function I started from
    $endgroup$
    – Begri
    Dec 11 '18 at 18:07










  • $begingroup$
    Yes, but for $f(t)mathrm{sign}(t)$ what $g$ do you take? It's quite obviuous... (And it's already spoiled in the answer...)
    $endgroup$
    – Federico
    Dec 11 '18 at 18:08




















  • $begingroup$
    What would be $g(t)$ in the Holder inequality?
    $endgroup$
    – Federico
    Dec 11 '18 at 18:01










  • $begingroup$
    I need to find $g(t)$ evaluating the function I started from
    $endgroup$
    – Begri
    Dec 11 '18 at 18:07










  • $begingroup$
    Yes, but for $f(t)mathrm{sign}(t)$ what $g$ do you take? It's quite obviuous... (And it's already spoiled in the answer...)
    $endgroup$
    – Federico
    Dec 11 '18 at 18:08


















$begingroup$
What would be $g(t)$ in the Holder inequality?
$endgroup$
– Federico
Dec 11 '18 at 18:01




$begingroup$
What would be $g(t)$ in the Holder inequality?
$endgroup$
– Federico
Dec 11 '18 at 18:01












$begingroup$
I need to find $g(t)$ evaluating the function I started from
$endgroup$
– Begri
Dec 11 '18 at 18:07




$begingroup$
I need to find $g(t)$ evaluating the function I started from
$endgroup$
– Begri
Dec 11 '18 at 18:07












$begingroup$
Yes, but for $f(t)mathrm{sign}(t)$ what $g$ do you take? It's quite obviuous... (And it's already spoiled in the answer...)
$endgroup$
– Federico
Dec 11 '18 at 18:08






$begingroup$
Yes, but for $f(t)mathrm{sign}(t)$ what $g$ do you take? It's quite obviuous... (And it's already spoiled in the answer...)
$endgroup$
– Federico
Dec 11 '18 at 18:08












1 Answer
1






active

oldest

votes


















0












$begingroup$

You can apply Hölder with $g=1$, i.e.
$$ left| int_{-1}^1 f(t) , text{sgn}(t) , mathrm dt right| leq int_{-1}^1 |f(t)|cdot 1 , mathrm dt leq left(int_{-1}^1 |f(t)|^2 , mathrm dtright)^{1/2}left(int_{-1}^1 1^2 , mathrm dtright)^{1/2} = sqrt 2left(int_{-1}^1 |f(t)|^2 , mathrm dtright)^{1/2}. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's very helpful for the OP, who didn't even give it a try
    $endgroup$
    – Federico
    Dec 11 '18 at 18:10










  • $begingroup$
    @Federico I read through the question too quickly, it seemed to me at first glance that he/she at least attempted to do something. My bad.
    $endgroup$
    – MisterRiemann
    Dec 11 '18 at 18:17













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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

You can apply Hölder with $g=1$, i.e.
$$ left| int_{-1}^1 f(t) , text{sgn}(t) , mathrm dt right| leq int_{-1}^1 |f(t)|cdot 1 , mathrm dt leq left(int_{-1}^1 |f(t)|^2 , mathrm dtright)^{1/2}left(int_{-1}^1 1^2 , mathrm dtright)^{1/2} = sqrt 2left(int_{-1}^1 |f(t)|^2 , mathrm dtright)^{1/2}. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's very helpful for the OP, who didn't even give it a try
    $endgroup$
    – Federico
    Dec 11 '18 at 18:10










  • $begingroup$
    @Federico I read through the question too quickly, it seemed to me at first glance that he/she at least attempted to do something. My bad.
    $endgroup$
    – MisterRiemann
    Dec 11 '18 at 18:17


















0












$begingroup$

You can apply Hölder with $g=1$, i.e.
$$ left| int_{-1}^1 f(t) , text{sgn}(t) , mathrm dt right| leq int_{-1}^1 |f(t)|cdot 1 , mathrm dt leq left(int_{-1}^1 |f(t)|^2 , mathrm dtright)^{1/2}left(int_{-1}^1 1^2 , mathrm dtright)^{1/2} = sqrt 2left(int_{-1}^1 |f(t)|^2 , mathrm dtright)^{1/2}. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's very helpful for the OP, who didn't even give it a try
    $endgroup$
    – Federico
    Dec 11 '18 at 18:10










  • $begingroup$
    @Federico I read through the question too quickly, it seemed to me at first glance that he/she at least attempted to do something. My bad.
    $endgroup$
    – MisterRiemann
    Dec 11 '18 at 18:17
















0












0








0





$begingroup$

You can apply Hölder with $g=1$, i.e.
$$ left| int_{-1}^1 f(t) , text{sgn}(t) , mathrm dt right| leq int_{-1}^1 |f(t)|cdot 1 , mathrm dt leq left(int_{-1}^1 |f(t)|^2 , mathrm dtright)^{1/2}left(int_{-1}^1 1^2 , mathrm dtright)^{1/2} = sqrt 2left(int_{-1}^1 |f(t)|^2 , mathrm dtright)^{1/2}. $$






share|cite|improve this answer









$endgroup$



You can apply Hölder with $g=1$, i.e.
$$ left| int_{-1}^1 f(t) , text{sgn}(t) , mathrm dt right| leq int_{-1}^1 |f(t)|cdot 1 , mathrm dt leq left(int_{-1}^1 |f(t)|^2 , mathrm dtright)^{1/2}left(int_{-1}^1 1^2 , mathrm dtright)^{1/2} = sqrt 2left(int_{-1}^1 |f(t)|^2 , mathrm dtright)^{1/2}. $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 18:01









MisterRiemannMisterRiemann

5,8171625




5,8171625












  • $begingroup$
    That's very helpful for the OP, who didn't even give it a try
    $endgroup$
    – Federico
    Dec 11 '18 at 18:10










  • $begingroup$
    @Federico I read through the question too quickly, it seemed to me at first glance that he/she at least attempted to do something. My bad.
    $endgroup$
    – MisterRiemann
    Dec 11 '18 at 18:17




















  • $begingroup$
    That's very helpful for the OP, who didn't even give it a try
    $endgroup$
    – Federico
    Dec 11 '18 at 18:10










  • $begingroup$
    @Federico I read through the question too quickly, it seemed to me at first glance that he/she at least attempted to do something. My bad.
    $endgroup$
    – MisterRiemann
    Dec 11 '18 at 18:17


















$begingroup$
That's very helpful for the OP, who didn't even give it a try
$endgroup$
– Federico
Dec 11 '18 at 18:10




$begingroup$
That's very helpful for the OP, who didn't even give it a try
$endgroup$
– Federico
Dec 11 '18 at 18:10












$begingroup$
@Federico I read through the question too quickly, it seemed to me at first glance that he/she at least attempted to do something. My bad.
$endgroup$
– MisterRiemann
Dec 11 '18 at 18:17






$begingroup$
@Federico I read through the question too quickly, it seemed to me at first glance that he/she at least attempted to do something. My bad.
$endgroup$
– MisterRiemann
Dec 11 '18 at 18:17




















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