= Solution
Differentiation satisfies $\|Tf\|_\infty\leq\|f\|$, so its <operator norm> is at most $1$. With $f_n(t)=\sin(2\pi nt)/(2\pi n)$, the ratio $\|Tf_n\|_\infty/\|f_n\|$ tends to $1$. Hence \b[$\|T\|=1$].
Normalize $g_n=f_n/(1+(2\pi n)^{-1})$, so $\|g_n\|=1$. Their derivatives are $\cos(2\pi nt)/(1+(2\pi n)^{-1})$. Distinct cosine functions are orthogonal in $L^2(0,1)$, and these normalizing factors stay bounded away from zero. Consequently the derivative images have pairwise $L^2$ distances bounded away from zero. Since <uniform convergence> would imply $L^2$ convergence, no subsequence converges uniformly. Thus \b[$T$ is not compact].
Back to article page
