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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 ?

Martin Sleziak
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Joe G.
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3 Answers3

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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.

Kelson Vieira
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  • 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! – QuantumDots Nov 19 '15 at 08:39
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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.

Andreas Blass
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    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 ? – Joe G. Nov 10 '12 at 03:43
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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.

Hans Engler
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    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). – Martin Argerami Nov 10 '12 at 05:18
  • @ 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." – Hans Engler Nov 10 '12 at 22:56
  • Yes indeed. It was a surprise for me too, as I was not aware of this property. – Martin Argerami Nov 10 '12 at 23:32