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.
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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 $\infty$. |
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No function can realize the upper bound. However, we can approach it by functions like $f_n(x):=\frac{\sin(2\pi n x)}n$. |
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