We had this exercise that uses Riesz' lemma to prove that a functional space actually has a finite dimension. I was curious to know what this space "looks" like, if we can find natural elements that would generate it. Here are the assumptions:

  • $E = \mathcal{C}^0([0,1], \mathbb{R}) $ is the space of continuous functions on $[0, 1]$, endowed with the norm $ ||f||_\infty = \sup\{|f(t)|, t \in [0, 1] \} $.
  • We have a closed linear subspace $F$ of $E$ such that $F \subset \mathcal{C}^1([0,1], \mathbb{R})$ (the space of continuously differentiable functions with continuous derivatives) and $$ \exists C > 0, \forall f \in F, \; ||f'||_\infty \leq C \, ||f||_\infty $$

You can show that the unit ball of that space is compact for $ ||\cdot||_\infty$ (using Ascoli), hence by Riesz' lemma the space has a finite dimension (!)

If I follow the conclusion then I imagine that you could build up a finite family of functions that would generate the whole space. Is that true? What would these functions look like? Are they useful somewhere else? Are they known?


2 Answers 2


Perhaps surprising but it is in fact possible to get an explicit bound for the dimension of $F$ (math magic at work). This is a bit like a constructive proof of Riesz's theorem.

We construct recursively subspaces of $F$: Let $f_1\in F$ and $x_1\in [0,1]$ be such that $f_1(x_1)=1=\|f_1\|_\infty$ and define $$ F_1 = \{ f\in F : f(x_1)=0 \}$$ If $F_1$ is non-trivial find $f_2\in F_1$ and $x_2\in [0,1]$ (necessarily different from $x_1$) such that $f_2(x_2)=1=\|f_2\|_\infty$. Then set $$ F_2=\{f\in F: f(x_1)=f(x_2)=0\}$$ Proceed inductively as long as you can. If finally $F_{d+1}=\{0\}$ your space is of dimension $d$. Now each $f_k$ vanishes at $x_i$, $i<k$ and $f_k(x_k)=1=\|f_k\|_\infty$ and by the MVT we must have $$ C\geq \|f_k'\|_\infty \geq \max \{ \frac{1}{|x_k-x_i|} : 1\leq i<k\}$$ Therefore, $$ |x_k-x_i| \geq 1/C$$ for all $1\leq i<k \leq d$. This implies $d\leq C+1$.

  • $\begingroup$ Interesting approach! This seems to be the way to relate the constant to the dimension of the space. So why would the induction end at all, why do we know there will be a $d$ such that $F_{d+1} = \{ 0 \}$? $\endgroup$ Aug 28, 2016 at 16:37
  • 2
    $\begingroup$ It is the last 'box' counting argument which I didn't detail. But if the distance between every couple of points is at least $1/C$ then you can not place more than $C+1$ of such points in the interval. $\endgroup$
    – H. H. Rugh
    Aug 28, 2016 at 16:44

I had this in the comments earlier, but when I went to add an example, I found it too frustrating to edit in the comment box.

Suppose that $C = 1$. So the condition of interest on a subspace $F$ is: $\|f'\| \leq \|f\|$ for all $f \in F$.

Since $(e^x)'=e^x$ and $(e^{-x})' = -e^{-x}$ it is easy to see that the lines $L_1 = \mathbb{R} \cdot e^x$ and $L_2 = \mathbb{R} \cdot e^{-x}$ both satisfy the condition. However, there cannot be a subspace $F$ satisfying the property which contains both $L_1$ and $L_2$. Such a space $F$ would need to contain $f = e^{x} - e^{-x}$, but one can check that $\|f'\| > \|f\|$, so this is impossible.

The conclusion from the above is that, for a given $C$, there is not just one example of a subspace $F$, but many. Moreover, there is not a largest one. That said, it might be interesting to ask if there is a finite bound on the dimensions of spaces satisfying the property for given $C$, and then search for examples realizing the maximal dimension.

  • $\begingroup$ It seems reasonable to expect that the number of dimensions would increase as $C$ gets larger then I guess? Coming up with these examples is the part I cannot fathom... $\endgroup$ Aug 28, 2016 at 16:05

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