If X = $C^1[0,1]$, $||f||_1 = ||f||_\infty$, and $||f||_2 = ||f||_\infty + ||f'||_\infty$, show that the identity map $I: (X, ||·||1) → (X, ||·||2)$ is bounded below, but not continuous.
I know how to show that it is bounded below ($m = 1$, since $||f||_\infty \leq ||f||_\infty + ||f'||_\infty$), but I'm a little stuck on showing it isn't continuous. My best guess is to show that it isn't bounded by contradiction, which would imply that it isn't continuous.
So I suppose that there exists some M so that $||If||_1 = ||f||_2 \leq M ||f||_1$, which through a little algebra tells me $||f'||_\infty \leq (M - 1)||f||_\infty$, and so $sup_{0 \leq x \leq 1} |lim_{h \rightarrow 0} {{f(x) - f(x + h)} \over {h}}| \leq (M-1) sup_{0 \leq x \leq 1} |f(x)|$, but I'm not sure how to reach a contradiction from here. Am I missing a step, or is there some other way of going about this?
Take a sequence $(f_n)_{n\in\Bbb N}\to f$ uniformly with $(f_n')_{n\in\Bbb N}\not\to f'$ uniformly. Try with $f_n'(x)= x^n$, $f_n(x)=\cdots$