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I'm trying to show that $D:(X, \|\cdot\|_\infty) \rightarrow C[0,1]$ is a continuous map. $D$ is the differential operator and $X$ is a closed (proper) subset of $C^1[0,1]$.

The fact that $X$ is closed in $C^1[0,1]$ must be important in the proof because otherwise this result is obviously false. However, I don't know how to use this fact.

I need this result to apply Arzela-Ascoli theorem to show that unit ball of X is compact and then conclude that $X$ is finite dimensional.

Does anyone know how to tackle this problem ?

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  • $\begingroup$ Is $D$ differential operator? $\endgroup$ – Kelson Vieira Nov 10 '12 at 3:04
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    $\begingroup$ Probably you meant to ask this: Differentiation continuous iff domain is finite dimensional $\endgroup$ – user48854 Nov 10 '12 at 3:18
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    $\begingroup$ Joe, I believe you want to show that: if $X\subset C^1[0,1]$ is closed in $(C[0,1], \|\cdot\|_{\infty})$, then $X$ is finite dimentional. Note that, $X$ is closed in $(C^1[0,1], \|\cdot\|_{C^1})$, where, $\|f\|_{C^1}:= \|f\|_{\infty} + \|f'\|_{\infty}$ (use the fundamental theorem of calculus and properties of uniform convergence). Apply open mapping theorem in $Id: (X, \|\cdot\|_{C^1}) \rightarrow (X, \|\cdot\|_{\infty})$ and show that the unit ball in $(X, \|\cdot\|_{C^1})$ is compact, by Arzela-Ascoli. Thus $X$ is finte dimentional! $\endgroup$ – Kelson Vieira Nov 10 '12 at 3:43
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    $\begingroup$ Thanks Kelson Vieira for clarifying the question and the answer ! $\endgroup$ – Joe G. Nov 10 '12 at 3:50
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    $\begingroup$ @KelsonVieira: I think you should write that as an answer, not a comment. $\endgroup$ – Martin Argerami Nov 10 '12 at 5:08
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First, note that, $X$ is closed in $(C^1[0,1],∥⋅∥_{C^1})$, where, $∥f∥_{C^1}:=∥f∥_{\infty}+∥Df∥_{\infty}$.

Indeed, suppose $(f_n)$ in $X$, such that $f_n \rightarrow f$ and $Df_n \rightarrow g$, uniformly. Then $f,g \in C[0,1]$ and, by fundamental theorem of calculus,

$$f_n(x)= f_n(0) + \int_0^x Df_n(t)dt.$$

So,

$$f(x)= f(0) + \int_0^x g(t)dt.$$ Thus, by fundamental theorem of calculus, $f\in C^1[0,1]$ and $Df=g$. This prove that $X$ is closed in $(C^1[0,1],∥⋅∥_{C^1})$.

Now, write $Id:(X,∥⋅∥_{C^11})→(X,∥⋅∥_{\infty})$. Note that $Id$ is a homeomorphism, By open mapping theorem. Thus, exist $c>0$, such that, for each $f\in X$, $$\|Df\|_{\infty} \leq \|f\|_{C^1}\leq c\|f\|_{\infty},$$ that is, $D:(X,\| \cdot\|_{\infty}) \longrightarrow C[0,1]$ is a continuous map.

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  • $\begingroup$ How does the open mapping theorem guarantee that inequality? And how does knowing the unit ball in $(X,||\cdot||_{C1})$ is compact allow us to conclude that $X$ is finite-dimensional? Thanks! $\endgroup$ – QuantumDots Nov 19 '15 at 8:39
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If $D$ is intended to be differentiation and if you really mean to use the infinity-norm (the supremum of the values of the function) on the domain, then $D$ isn't continuous. You can have functions in $C^1$ that are very small in the infinity-norm but whose derivatives get very big --- a small function that wiggles a lot, like $\frac1n\sin(n^{100}x)$.

Perhaps the norm on the domain of $D$ should have been a more typical norm on $C^1$, taking into account not only the magnitude of the function but also the magnitude of its derivative. Then $D$ would be continuous, and the proof of that would be very easy.

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    $\begingroup$ I forgot to mention that $D$ is indeed the differential operator. I'm interested in the case where $X$ a proper closed subspace of $C^1[0,1]$. Could $D$ be continuous in this case ? $\endgroup$ – Joe G. Nov 10 '12 at 3:43
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Take $X = C^1[0,1]$ and let $f_n \in X$ be $f_n(x) = \frac{1}{\sqrt{n}}x^n$. Then $\|f_n\|_{\infty} = \frac{1}{\sqrt{n}}$, but $\|Df_n\|_{\infty} = \sqrt{n}$. Hence $D$ is not continuous.

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    $\begingroup$ Although $C^1[0,1]$ is tautologically closed, it is not complete. If you notice, the question asks for $X$ to be proper. And in that situation, $D$ is indeed continuous (because $X$ is finite-dimensional). $\endgroup$ – Martin Argerami Nov 10 '12 at 5:18
  • $\begingroup$ @ Martin - good point :) I had overlooked this condition. The statement to prove is really "any proper subspace of $C^1$ that is closed under the $\|\cdot\|_{\infty}$ norm is finite dimensional." $\endgroup$ – Hans Engler Nov 10 '12 at 22:56
  • $\begingroup$ Yes indeed. It was a surprise for me too, as I was not aware of this property. $\endgroup$ – Martin Argerami Nov 10 '12 at 23:32

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