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Let $A\subset C([0,1])$ a closed linear subspace with respect to the usual supremum norm satisfying $A\subset C^1([0,1])$. Is $D\colon A\rightarrow C([0,1]), \ f\rightarrow f'$ continuous iff $A$ is finite dimensional?

If $A$ is finite dimensional $D$ is continuous of course. But is the other implication true at all? Wouldn't something like $A:=\overline{\text{span}\{\sin{\left(t+\frac{1}{n}\right)} \ | \ n\in\mathbb{N}\}}$ be a counterexample?

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    Yes boundedness is clear. I don't see $h$ow to prove equicontinuity. the unit ball is compact iff the space is finite dimensional2012-07-31

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From the dicussion above it follows that it is enough to show equicontinuiuty of $\mathrm{Ball}_A(0,1)$ provided $D$ is continuous.

Fix $\varepsilon>0$ and take $\delta=(2\Vert D\Vert)^{-1}\varepsilon$. From mean value theorem it follows that for all $x_1,x_2\in [0,1]$ such that $|x_2-x_1|\leq\delta$ and all $f\in\mathrm{Ball}_A(0,1)$ we have $ |f(x_2)-f(x_1)|=|f'(\xi)(x_2-x_1)|=|f'(\xi)||x_2-x_1|\leq\Vert f'\Vert_\infty\delta\leq $ $ \Vert D\Vert\Vert f\Vert_\infty\delta\leq\Vert D\Vert\cdot 1\cdot \frac{\varepsilon}{2\Vert D\Vert}<\varepsilon $ This means that $\mathrm{Ball}_A(0,1)$ is equicontinuous.

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    The reason why the intended counterexample (the closed span of shifted sines) doesn't work is that it is not contained in $C^1$. It's a good exercise to find out why.2012-07-31