Show that the functional $F$ on $L_p(-1,1)$ given by $ F(f) = int_{-1}^0 f(t)dt - int_{0}^1 f(t)dt $ is...












1












$begingroup$


I have to show that the functional $F$: $$ F(f) = int_{-1}^0 f(t)dt - int_{0}^1 f(t)dt $$



on the space $L_p(-1, 1), pgeqslant 1 $, is continuous.



Solution:



I need to get an inequality $$ vert F(f) vert le M left(int_{-1}^1 bigvert f(t)big vert ^p dtright)^{1/p}.$$



Here $M$ is a constant and $M>0$.



I started:



$$bigvert F(f) bigvert=leftvert int_{-1}^0 f(t)dt - int_{0}^1 f(t)dt rightvert .$$



And here I stopped. Next I have to evaluate these two integrals by new $int_{-1}^1 g(t)dt$ function and here I have a problem with this. The next step is to use the Hölder's inequality to get the inequality I need to show (with $M>0$).



Question:



How to find $int_{-1}^1 g(t)dt$?






functional-analysis functions inequality continuity lp-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Is there a reason to write $$F(f)=int_{-1}^0,f(t),text{d}t+int_0^1,f(t),text{d}t$$ instead of just $$F(f)=int_{-1}^1,f(t),text{d}t,?$$ (That is, did you make any typo?)
    $endgroup$
    – Batominovski
    Dec 11 '18 at 20:49










  • $begingroup$
    Oh.. sorry, $-$ have to be there
    $endgroup$
    – Philip
    Dec 11 '18 at 20:52
















1












$begingroup$


I have to show that the functional $F$: $$ F(f) = int_{-1}^0 f(t)dt - int_{0}^1 f(t)dt $$



on the space $L_p(-1, 1), pgeqslant 1 $, is continuous.



Solution:



I need to get an inequality $$ vert F(f) vert le M left(int_{-1}^1 bigvert f(t)big vert ^p dtright)^{1/p}.$$



Here $M$ is a constant and $M>0$.



I started:



$$bigvert F(f) bigvert=leftvert int_{-1}^0 f(t)dt - int_{0}^1 f(t)dt rightvert .$$



And here I stopped. Next I have to evaluate these two integrals by new $int_{-1}^1 g(t)dt$ function and here I have a problem with this. The next step is to use the Hölder's inequality to get the inequality I need to show (with $M>0$).



Question:



How to find $int_{-1}^1 g(t)dt$?






functional-analysis functions inequality continuity lp-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Is there a reason to write $$F(f)=int_{-1}^0,f(t),text{d}t+int_0^1,f(t),text{d}t$$ instead of just $$F(f)=int_{-1}^1,f(t),text{d}t,?$$ (That is, did you make any typo?)
    $endgroup$
    – Batominovski
    Dec 11 '18 at 20:49










  • $begingroup$
    Oh.. sorry, $-$ have to be there
    $endgroup$
    – Philip
    Dec 11 '18 at 20:52














1












1








1


1



$begingroup$


I have to show that the functional $F$: $$ F(f) = int_{-1}^0 f(t)dt - int_{0}^1 f(t)dt $$



on the space $L_p(-1, 1), pgeqslant 1 $, is continuous.



Solution:



I need to get an inequality $$ vert F(f) vert le M left(int_{-1}^1 bigvert f(t)big vert ^p dtright)^{1/p}.$$



Here $M$ is a constant and $M>0$.



I started:



$$bigvert F(f) bigvert=leftvert int_{-1}^0 f(t)dt - int_{0}^1 f(t)dt rightvert .$$



And here I stopped. Next I have to evaluate these two integrals by new $int_{-1}^1 g(t)dt$ function and here I have a problem with this. The next step is to use the Hölder's inequality to get the inequality I need to show (with $M>0$).



Question:



How to find $int_{-1}^1 g(t)dt$?






functional-analysis functions inequality continuity lp-spaces







share|cite|improve this question











$endgroup$




I have to show that the functional $F$: $$ F(f) = int_{-1}^0 f(t)dt - int_{0}^1 f(t)dt $$



on the space $L_p(-1, 1), pgeqslant 1 $, is continuous.



Solution:



I need to get an inequality $$ vert F(f) vert le M left(int_{-1}^1 bigvert f(t)big vert ^p dtright)^{1/p}.$$



Here $M$ is a constant and $M>0$.



I started:



$$bigvert F(f) bigvert=leftvert int_{-1}^0 f(t)dt - int_{0}^1 f(t)dt rightvert .$$



And here I stopped. Next I have to evaluate these two integrals by new $int_{-1}^1 g(t)dt$ function and here I have a problem with this. The next step is to use the Hölder's inequality to get the inequality I need to show (with $M>0$).



Question:



How to find $int_{-1}^1 g(t)dt$?






functional-analysis functions inequality continuity lp-spaces




functional-analysis functions inequality continuity lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 21:27









Batominovski

33.1k33293




33.1k33293










asked Dec 11 '18 at 20:44









PhilipPhilip

857




857








  • 2




    $begingroup$
    Is there a reason to write $$F(f)=int_{-1}^0,f(t),text{d}t+int_0^1,f(t),text{d}t$$ instead of just $$F(f)=int_{-1}^1,f(t),text{d}t,?$$ (That is, did you make any typo?)
    $endgroup$
    – Batominovski
    Dec 11 '18 at 20:49










  • $begingroup$
    Oh.. sorry, $-$ have to be there
    $endgroup$
    – Philip
    Dec 11 '18 at 20:52














  • 2




    $begingroup$
    Is there a reason to write $$F(f)=int_{-1}^0,f(t),text{d}t+int_0^1,f(t),text{d}t$$ instead of just $$F(f)=int_{-1}^1,f(t),text{d}t,?$$ (That is, did you make any typo?)
    $endgroup$
    – Batominovski
    Dec 11 '18 at 20:49










  • $begingroup$
    Oh.. sorry, $-$ have to be there
    $endgroup$
    – Philip
    Dec 11 '18 at 20:52








2




2




$begingroup$
Is there a reason to write $$F(f)=int_{-1}^0,f(t),text{d}t+int_0^1,f(t),text{d}t$$ instead of just $$F(f)=int_{-1}^1,f(t),text{d}t,?$$ (That is, did you make any typo?)
$endgroup$
– Batominovski
Dec 11 '18 at 20:49




$begingroup$
Is there a reason to write $$F(f)=int_{-1}^0,f(t),text{d}t+int_0^1,f(t),text{d}t$$ instead of just $$F(f)=int_{-1}^1,f(t),text{d}t,?$$ (That is, did you make any typo?)
$endgroup$
– Batominovski
Dec 11 '18 at 20:49












$begingroup$
Oh.. sorry, $-$ have to be there
$endgroup$
– Philip
Dec 11 '18 at 20:52




$begingroup$
Oh.. sorry, $-$ have to be there
$endgroup$
– Philip
Dec 11 '18 at 20:52










1 Answer
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$begingroup$

Using the Triangle Inequality, we have
$$big|F(f)big|leq int_{-1}^{+1},big|f(x)big|,text{d}xtext{ for each }fin L^pbig((-1,+1)big),.$$
By Hölder's Inequality,
$$begin{align}int_{-1}^{+1},big|f(x)big|,text{d}x&=int_{-1}^{+1},big|f(x)big|cdot 1,text{d}x\&leq left(int_{-1}^{+1},big|f(x)big|^p,text{d}xright)^{frac{1}{p}},left(int_{-1}^{+1},1^q,text{d}xright)^{frac{1}{q}}text{ for each }fin L^pbig((-1,+1)big),,end{align}$$
where $qin[1,infty]$ is such that $dfrac1p+dfrac1q=1$. This shows that
$$big|F(f)big|leq |f|_p,2^{frac{1}{q}}=2^{1-frac1p},|f|_ptext{ for each }fin L^pbig((-1,+1)big),,$$
or
$$|F|_{L^p_text{op}}leq 2^{1-frac1p},.$$
By taking $f:(-1,+1)tomathbb{C}$ to be the step function
$$f(x)=begin{cases}-1&text{if }-1<x<0,,\+1&text{if }0leq x <+1,,end{cases}$$
we can see that the operator norm $|F|_{L^p_text{op}}$ of $F$ does indeed equal $2^{1-frac1p}$ (thus, $F$ is a bounded linear functional, whence continuous). With a little tweak, this result is also true for $p=infty$, not just $pin[1,infty)$.






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    $begingroup$

    Using the Triangle Inequality, we have
    $$big|F(f)big|leq int_{-1}^{+1},big|f(x)big|,text{d}xtext{ for each }fin L^pbig((-1,+1)big),.$$
    By Hölder's Inequality,
    $$begin{align}int_{-1}^{+1},big|f(x)big|,text{d}x&=int_{-1}^{+1},big|f(x)big|cdot 1,text{d}x\&leq left(int_{-1}^{+1},big|f(x)big|^p,text{d}xright)^{frac{1}{p}},left(int_{-1}^{+1},1^q,text{d}xright)^{frac{1}{q}}text{ for each }fin L^pbig((-1,+1)big),,end{align}$$
    where $qin[1,infty]$ is such that $dfrac1p+dfrac1q=1$. This shows that
    $$big|F(f)big|leq |f|_p,2^{frac{1}{q}}=2^{1-frac1p},|f|_ptext{ for each }fin L^pbig((-1,+1)big),,$$
    or
    $$|F|_{L^p_text{op}}leq 2^{1-frac1p},.$$
    By taking $f:(-1,+1)tomathbb{C}$ to be the step function
    $$f(x)=begin{cases}-1&text{if }-1<x<0,,\+1&text{if }0leq x <+1,,end{cases}$$
    we can see that the operator norm $|F|_{L^p_text{op}}$ of $F$ does indeed equal $2^{1-frac1p}$ (thus, $F$ is a bounded linear functional, whence continuous). With a little tweak, this result is also true for $p=infty$, not just $pin[1,infty)$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Using the Triangle Inequality, we have
      $$big|F(f)big|leq int_{-1}^{+1},big|f(x)big|,text{d}xtext{ for each }fin L^pbig((-1,+1)big),.$$
      By Hölder's Inequality,
      $$begin{align}int_{-1}^{+1},big|f(x)big|,text{d}x&=int_{-1}^{+1},big|f(x)big|cdot 1,text{d}x\&leq left(int_{-1}^{+1},big|f(x)big|^p,text{d}xright)^{frac{1}{p}},left(int_{-1}^{+1},1^q,text{d}xright)^{frac{1}{q}}text{ for each }fin L^pbig((-1,+1)big),,end{align}$$
      where $qin[1,infty]$ is such that $dfrac1p+dfrac1q=1$. This shows that
      $$big|F(f)big|leq |f|_p,2^{frac{1}{q}}=2^{1-frac1p},|f|_ptext{ for each }fin L^pbig((-1,+1)big),,$$
      or
      $$|F|_{L^p_text{op}}leq 2^{1-frac1p},.$$
      By taking $f:(-1,+1)tomathbb{C}$ to be the step function
      $$f(x)=begin{cases}-1&text{if }-1<x<0,,\+1&text{if }0leq x <+1,,end{cases}$$
      we can see that the operator norm $|F|_{L^p_text{op}}$ of $F$ does indeed equal $2^{1-frac1p}$ (thus, $F$ is a bounded linear functional, whence continuous). With a little tweak, this result is also true for $p=infty$, not just $pin[1,infty)$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Using the Triangle Inequality, we have
        $$big|F(f)big|leq int_{-1}^{+1},big|f(x)big|,text{d}xtext{ for each }fin L^pbig((-1,+1)big),.$$
        By Hölder's Inequality,
        $$begin{align}int_{-1}^{+1},big|f(x)big|,text{d}x&=int_{-1}^{+1},big|f(x)big|cdot 1,text{d}x\&leq left(int_{-1}^{+1},big|f(x)big|^p,text{d}xright)^{frac{1}{p}},left(int_{-1}^{+1},1^q,text{d}xright)^{frac{1}{q}}text{ for each }fin L^pbig((-1,+1)big),,end{align}$$
        where $qin[1,infty]$ is such that $dfrac1p+dfrac1q=1$. This shows that
        $$big|F(f)big|leq |f|_p,2^{frac{1}{q}}=2^{1-frac1p},|f|_ptext{ for each }fin L^pbig((-1,+1)big),,$$
        or
        $$|F|_{L^p_text{op}}leq 2^{1-frac1p},.$$
        By taking $f:(-1,+1)tomathbb{C}$ to be the step function
        $$f(x)=begin{cases}-1&text{if }-1<x<0,,\+1&text{if }0leq x <+1,,end{cases}$$
        we can see that the operator norm $|F|_{L^p_text{op}}$ of $F$ does indeed equal $2^{1-frac1p}$ (thus, $F$ is a bounded linear functional, whence continuous). With a little tweak, this result is also true for $p=infty$, not just $pin[1,infty)$.






        share|cite|improve this answer











        $endgroup$



        Using the Triangle Inequality, we have
        $$big|F(f)big|leq int_{-1}^{+1},big|f(x)big|,text{d}xtext{ for each }fin L^pbig((-1,+1)big),.$$
        By Hölder's Inequality,
        $$begin{align}int_{-1}^{+1},big|f(x)big|,text{d}x&=int_{-1}^{+1},big|f(x)big|cdot 1,text{d}x\&leq left(int_{-1}^{+1},big|f(x)big|^p,text{d}xright)^{frac{1}{p}},left(int_{-1}^{+1},1^q,text{d}xright)^{frac{1}{q}}text{ for each }fin L^pbig((-1,+1)big),,end{align}$$
        where $qin[1,infty]$ is such that $dfrac1p+dfrac1q=1$. This shows that
        $$big|F(f)big|leq |f|_p,2^{frac{1}{q}}=2^{1-frac1p},|f|_ptext{ for each }fin L^pbig((-1,+1)big),,$$
        or
        $$|F|_{L^p_text{op}}leq 2^{1-frac1p},.$$
        By taking $f:(-1,+1)tomathbb{C}$ to be the step function
        $$f(x)=begin{cases}-1&text{if }-1<x<0,,\+1&text{if }0leq x <+1,,end{cases}$$
        we can see that the operator norm $|F|_{L^p_text{op}}$ of $F$ does indeed equal $2^{1-frac1p}$ (thus, $F$ is a bounded linear functional, whence continuous). With a little tweak, this result is also true for $p=infty$, not just $pin[1,infty)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 12 '18 at 0:55

























        answered Dec 11 '18 at 21:14









        BatominovskiBatominovski

        33.1k33293




        33.1k33293






























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