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.


2 Answers 2


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.

  • $\begingroup$ $C[0,1]$ is not a hilbert space, how do we get orthogonality of functions? $\endgroup$
    – yoshi
    Jan 18, 2019 at 16:03
  • $\begingroup$ This orthonormal basis is defined with respect to the inner product $\int_0^1 fg\,dx$. $\endgroup$ Jan 18, 2019 at 17:36
  • $\begingroup$ $C[0,1]$ doesnt have an inner product though right? math.stackexchange.com/questions/2669051/… $\endgroup$
    – yoshi
    Jan 18, 2019 at 18:00
  • 2
    $\begingroup$ It doesn't have an inner product, but several inner products can be defined in $C[0,1]$. $\endgroup$ Jan 20, 2019 at 10:23
  • $\begingroup$ @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. $\endgroup$
    – Eparoh
    Jan 8, 2022 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.


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