Show that the operator $T$ is continuous given that whenever $f(x)\geq 0$ for all $x\in[0,1]$, it follows that $(Tf)(x)\geq 0$ for all $x\in [0,1]$

Question

Suppose a linear operator $T:(C([0,1],\mathbb R),\lvert\lvert\cdot\rvert\rvert_{\infty})\to (C([0,1],\mathbb R),\lvert\lvert\cdot\rvert\rvert_{\infty})$ is such that whenever $f(x)\geq 0$ for all $x\in[0,1]$, it follows that $(Tf)(x)\geq 0$ for all $x\in [0,1]$. Show that $T$ is continuous and show that the operator norm of $T$ is $\lvert\lvert T1\rvert\rvert_{\infty}$.

I've tried showing that $T$ is bounded and hence continuous but it's hard to do without knowing explicitly what $T$ is. It's clear from the second part of the question that I should be able to show $\lvert\lvert Tf\rvert\rvert_{\infty}\leq\lvert\lvert T1\rvert\rvert_{\infty}\lvert\lvert f\rvert\rvert_{\infty}$ for all $f\in C[0,1]$ but I can't. I'm not sure how to use the property of $T$ to show continuity directly. I've tried showing continuity at $0$ but haven't made any progress, this problem seems related to real analysis which I haven't studied in a while.

Answer 1

Take $f \in C[0,1]$ and consider $g \in C[0,1]$ defined by $g(x) = \|f\|_{\infty} - f(x)$ for $x \in [0,1]$. Since $g(x) \ge 0$ for all $x \in [0,1]$, we have $$ (Tg)(x) \ge 0 \,\,\,\, \implies \,\,\,\, (T[\|f\|_\infty - f])(x) \ge 0 \,\,\,\, \implies \,\,\,\, \|f\|_\infty (T1)(x) - (Tf)(x) \ge 0,$$ for all $x \in [0,1].$ Rearranging and then passing to the supremum yields $$(Tf)(x) \le (T1)(x) \|f\|_\infty \,\,\,\,\, \implies \,\,\,\,\,\, \|Tf\|_\infty \le \|T1\|_\infty \|f\|_\infty.$$ And of course, equality holds when you plug in $f \equiv 1$. This shows that $T$ is bounded with operator norm $\|T1\|_{\infty}.$

EDIT: I suppose this is slightly incomplete. You should do the same steps with $h = \|f\|_\infty + f$ to arrive at $$-(Tf)(x) \le (T1)(x)\|f\|_\infty$$ so that (after combining with the other inequality) you have $$\lvert (Tf)(x) \rvert \le (T1)(x)\|f\|_\infty$$ before passing to the supremum