$T$ is bounded linear operator












0












$begingroup$


Question:
Let $T:C[0,1] to C[0,1]$, where $T(f) = (t^{2} +2) f(t)$.
Then $T$ is bounded linear operator.



Is this a true or a false statement? Justify your answer.



My solution:
$$
| T(f) | = | (t^{2} +2) f(t) | leq |(t^{2} +2)| |f(t)|
leq 3 |f|
$$
which is finite since it's bounded



Is it true?



Thanks a lot.






functional-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that boundedness is not the only condition you have to check, also linearity ofcourse.
    $endgroup$
    – Bo5man
    Dec 19 '18 at 13:37
















0












$begingroup$


Question:
Let $T:C[0,1] to C[0,1]$, where $T(f) = (t^{2} +2) f(t)$.
Then $T$ is bounded linear operator.



Is this a true or a false statement? Justify your answer.



My solution:
$$
| T(f) | = | (t^{2} +2) f(t) | leq |(t^{2} +2)| |f(t)|
leq 3 |f|
$$
which is finite since it's bounded



Is it true?



Thanks a lot.






functional-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that boundedness is not the only condition you have to check, also linearity ofcourse.
    $endgroup$
    – Bo5man
    Dec 19 '18 at 13:37














0












0








0





$begingroup$


Question:
Let $T:C[0,1] to C[0,1]$, where $T(f) = (t^{2} +2) f(t)$.
Then $T$ is bounded linear operator.



Is this a true or a false statement? Justify your answer.



My solution:
$$
| T(f) | = | (t^{2} +2) f(t) | leq |(t^{2} +2)| |f(t)|
leq 3 |f|
$$
which is finite since it's bounded



Is it true?



Thanks a lot.






functional-analysis







share|cite|improve this question











$endgroup$




Question:
Let $T:C[0,1] to C[0,1]$, where $T(f) = (t^{2} +2) f(t)$.
Then $T$ is bounded linear operator.



Is this a true or a false statement? Justify your answer.



My solution:
$$
| T(f) | = | (t^{2} +2) f(t) | leq |(t^{2} +2)| |f(t)|
leq 3 |f|
$$
which is finite since it's bounded



Is it true?



Thanks a lot.






functional-analysis




functional-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 14:50









YukiJ

2,1202928




2,1202928










asked Dec 19 '18 at 13:16









Duaa HamzehDuaa Hamzeh

524




524








  • 1




    $begingroup$
    Note that boundedness is not the only condition you have to check, also linearity ofcourse.
    $endgroup$
    – Bo5man
    Dec 19 '18 at 13:37














  • 1




    $begingroup$
    Note that boundedness is not the only condition you have to check, also linearity ofcourse.
    $endgroup$
    – Bo5man
    Dec 19 '18 at 13:37








1




1




$begingroup$
Note that boundedness is not the only condition you have to check, also linearity ofcourse.
$endgroup$
– Bo5man
Dec 19 '18 at 13:37




$begingroup$
Note that boundedness is not the only condition you have to check, also linearity ofcourse.
$endgroup$
– Bo5man
Dec 19 '18 at 13:37










1 Answer
1






active

oldest

votes


















1












$begingroup$

I think that $C[0,1]$ is equipped with the norm $||f||:= max {|f(t)|: t in [0,1]}.$



Your solution is correct, but not nicely written.



For $t in [0,1]$ and $f in C[0,1]$ we have



$|T(f)(t)|=(t^2+2)|f(t)| le 3 |f(t)| le 3 ||f||.$



Thus



$||T(f)|| le 3 ||f||$, which shows that $T$ is bounded.






share|cite|improve this answer









$endgroup$













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    1 Answer
    1






    active

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    active

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    active

    oldest

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    1












    $begingroup$

    I think that $C[0,1]$ is equipped with the norm $||f||:= max {|f(t)|: t in [0,1]}.$



    Your solution is correct, but not nicely written.



    For $t in [0,1]$ and $f in C[0,1]$ we have



    $|T(f)(t)|=(t^2+2)|f(t)| le 3 |f(t)| le 3 ||f||.$



    Thus



    $||T(f)|| le 3 ||f||$, which shows that $T$ is bounded.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I think that $C[0,1]$ is equipped with the norm $||f||:= max {|f(t)|: t in [0,1]}.$



      Your solution is correct, but not nicely written.



      For $t in [0,1]$ and $f in C[0,1]$ we have



      $|T(f)(t)|=(t^2+2)|f(t)| le 3 |f(t)| le 3 ||f||.$



      Thus



      $||T(f)|| le 3 ||f||$, which shows that $T$ is bounded.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I think that $C[0,1]$ is equipped with the norm $||f||:= max {|f(t)|: t in [0,1]}.$



        Your solution is correct, but not nicely written.



        For $t in [0,1]$ and $f in C[0,1]$ we have



        $|T(f)(t)|=(t^2+2)|f(t)| le 3 |f(t)| le 3 ||f||.$



        Thus



        $||T(f)|| le 3 ||f||$, which shows that $T$ is bounded.






        share|cite|improve this answer









        $endgroup$



        I think that $C[0,1]$ is equipped with the norm $||f||:= max {|f(t)|: t in [0,1]}.$



        Your solution is correct, but not nicely written.



        For $t in [0,1]$ and $f in C[0,1]$ we have



        $|T(f)(t)|=(t^2+2)|f(t)| le 3 |f(t)| le 3 ||f||.$



        Thus



        $||T(f)|| le 3 ||f||$, which shows that $T$ is bounded.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 13:25









        FredFred

        48.6k11849




        48.6k11849






























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