Consider the operator $T: C^2[0,1] \subset C^1[0,1] \to C^1[0,1]$ defined by $Tf=f'+f''$. Compute $\| T e^{-nx } \|_{\infty }$ and $\| T x^n \|_{\infty }$.
My attempt.
First I tried to compute $\| e^{-nx } \|_{\infty } $ and $\| x^n \|_{\infty } $. We have $$\| e^{-nx } \|_{\infty } =\sup_{x \in \mathbb{R} } |e^{-nx} |$$ We find the derivative of $|e^{-nx} |$ which is $-ne^{-nx} $, and $$-ne^{-nx} =0 \implies n=0$$ So $$\| e^{-nx } \|_{\infty }=1$$
Similarly for $$\| x^n \|_{\infty } =\sup_{x \in \mathbb{R} } |x^n |$$ So the derivative is $|nx^{n-1} |$ which is $0$ for $n=0$ Hence, $$\| x^n \|_{\infty }=1.$$
Now for $\| T e^{-nx } \|_{\infty }$, we have $$\| Te^{-nx} \|_{\infty } =\| -ne^{-nx} +n^2 e^{-nx} \|_{\infty } =\| (n(n-1)) e^{-nx} \|_{\infty } =-n+n^2$$ therefore $\| e^{-nx } \|_{\infty } =-n+n^2$.
For $\| T x^n \|_{\infty }$ we have $$\| Tx^n \|_{\infty } =\| nx^{n-1} +n(n-1)x^{n-2} \| \leq n \| x^{n-1} \|_{\infty } +n(n-1) \| x^{n-2} \|_{\infty }=n+n(n-1).$$
Is this correct? I think there's something wrong with my computation for $\|Tx^n \|_{\infty }$ because if I used the derivative to find Maxima of $| T x^n |$ I would get $| n(n-1) x^{n-2}-n(n-1)(n-2) | =0$ and it seems that either $n=0$ or $n=1$ or $x=0$ or $x=n-2$, in each case I get different extrema.
Can you explain what is wrong with my computation?