# 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 $x\in [0,1]$

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.

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