Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider $T:C^1[0, 1]\rightarrow C[0, 1]$ given by $Tf=f'$ where $$\|f\|_{C^1}=\|f\|_\infty+\|f'\|_\infty$$ and $\|f\|_\infty=\sup_{x\in [0, 1]}|f(x)|$. How to prove $\|T\|=1$? The inequality $\|T\|\leq 1$ is easy, but I can't find a function for showing this upper bound is achieved.

share|improve this question
    
how can we show above equality ||f||₁=||f′||_{∞}+||f||_{∞} –  1214 May 3 at 13:25

2 Answers 2

up vote 3 down vote accepted

Consider the functions $$ f_n(x)=x^n $$ for $n\geq 1$. Then $$ \|f_n\|_{C^1}=\|x^n\|_\infty+\|nx^{n-1}\|_\infty =n+1 $$ and $$ \|Tf_n\|_\infty=\|nx^{n-1}\|_\infty=n. $$ So $$ \|T\|\geq \frac{n}{n+1} $$ for all $n$.

Hence $\|T\|\geq 1$ when letting $n$ tend to $\infty$.

share|improve this answer
    
@DavidMitra Oh! Thanks! What a trick. –  1015 Feb 16 '13 at 20:58
    
Nice Proof, thanks =D –  PtF Feb 17 '13 at 11:13

No function can realize the upper bound.

However, we can approach it by functions like $f_n(x):=\frac{\sin(2\pi n x)}n$.

share|improve this answer
    
I was almost sure that approximation would be needed...Thanks for the tip.. –  PtF Feb 16 '13 at 20:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.