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.

  • 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

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

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


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


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