# Finite dimensional subspace of $C([0,1])$

Let linear $S$ be a subspace of $C([0,1])$, i.e., the continuous real-valued functions on $[0,1]$. Assume that there exists $c>0$, such that $\|\,f\|_\infty\leq c \|\,f\|_2$, for all $f\in S$. Then show that $S$ is finite-dimensional.

This is equivalent to proving that the closed unit ball in $(S,\|\|_\infty)$ or in $(S,\|\|_2)$ is compact, but I can't derive this.

Another thought is that the $L^2$ closure of $S$ is a subset of $C([0,1])$. Indeed, if $f_n\to f\in L^2$, then the given relation $\|f_n-f\|_\infty \leq c\|f_n-f\|_2$ implies that $f_n\to f$ in $L^\infty$, so $f$ is continuous. Therefore we have the inclusion $$S\subset \overline{S}^{L^2}\subset C([0,1])\subset L^2$$ If $S$ was infinite dimensional, then $\overline{S}$ would be an infinite dimensional hilbert space, proper subset of $L^2$.

Another thought is that all the $L^p$ norms on $S$ are equivalent (by Holder). If ALL norms are equivalent, then $S$ has to be finite dimensional.

Let me provide a detailed proof:

Assume that $$\{v_1,\ldots, v_n\}\subset S$$ is an orthonormal set functions in $$L^2[0,1]$$, i.e., $$\int_0^1 v_iv_j\,dx=\delta_{ij}$$. For a fixed $$a=(a_1,\ldots,a_n)\in\mathbb R^n$$ we define the mapping $$\varPhi_a :\mathbb R^n\to \mathbb R$$ as $$\varPhi_a(x)=\sum_{j=1}^n a_jv_j(x).$$ Then $$\|\varPhi_a\|_{L^2}^2=\int_0^2 \varPhi_a^2(x)\,dx= \sum_{j,k=1}^n\int_0^1a_ja_kv_jv_k \,dx=\sum_{j=1}^n a_j^2=\|a\|^2,$$ and thus, for every $$x\in [0,1]$$ $$\lvert a_1v_1(x)+\cdots+a_nv_n(x)\rvert =\lvert\varPhi_a(x)\rvert \le \|\varPhi_a\|_{L^\infty} \le c\|\varPhi_a\|_{L^2}\le c\|a\|,$$ and as the above holds for every $$a\in\mathbb R^n$$, we conclude that $$v_1^2(x)+\cdots+v_n^2(x)\le c^2,\tag{1}$$ for every $$x\in [0,1]$$. Here we have used the fact that, if $$\sum_{j=1}^n a_jb_j\le c\Big(\sum_{j=1}^n a_j^2\Big)^{1/2},$$ for all $$a_1,\ldots,a_n$$, then setting $$a_j=b_j$$, $$j=1,\ldots,n$$, we obtain that $$\sum_{j=1}^n b_j^2\le c^2$$.

Integrating $$(1)$$ over $$[0,1]$$ we obtain that $$n\le c^2.$$ We have derived that $$\,\dim S \le c^2$$.

Note. All the above can be easily generalized in the complex case.

• $C[0,1]$ is not a hilbert space, how do we get orthogonality of functions? Jan 18, 2019 at 16:03
• This orthonormal basis is defined with respect to the inner product $\int_0^1 fg\,dx$. Jan 18, 2019 at 17:36
• $C[0,1]$ doesnt have an inner product though right? math.stackexchange.com/questions/2669051/… Jan 18, 2019 at 18:00
• It doesn't have an inner product, but several inner products can be defined in $C[0,1]$. Jan 20, 2019 at 10:23
• @yoshi $C[0,1]$ with the inner product induced by $L^2$ is not complete, but an inner space. So, taking any $n$ lineraly independent vectors in $S$, you can apply Gram-Schmidt to obtain $n$ orthonormal vectors respect to the induced inner product. From there, you can just follow Yiorgos proof to conclude that if that $n$ exists then it is less than $c^2$ and so $S$ cannot be infinite dimensional. Jan 8 at 10:31

You can apply a theorem of Grothendieck to the closure of $S$ in $L^2$ which is (as you show) contained in $C([0,1]) \subseteq L^\infty$.

Grothendieck's theorem says that every closed subspace of $L^p(\mu)$ (where $\mu$ is a probability measure on some measurable space and $0<p<\infty$) which is contained in $L^\infty(\mu)$ is finite dimensional.

A convenient reference is Rudin's Functional Analysis, Theorem 5.2.