I found the following question in an exam and not able to completely crack it. I paste the question as it was there:

Suppose $C[a, b]$ denote the set of all real valued continuous functions on $[a, b]$. Define an inner product on it so that it will become an inner product space having the property that for any sequence $\{f_n\}_{n=1}^{\infty} \subset C[a, b] ~ ||f_n-f_m|| \to 0$ as $m, n \to \infty$ if and only if for some $f, ||f_n-f||\to 0$ as $n \to \infty$.

What I understand that this question is asking for an 'inner product' on $C[a, b]$ so that, a sequence $(f_n)$ is Cauchy iff it converges ! i.e., to define an 'complete inner product' on the vector space $C[a, b]$.

  • Now although, $\langle f, g \rangle :=\int_a^bfg$ defines an inner product on $C[a, b]$, it is not complete! So we have to define something else.
  • Also another fact is that the (norm)topology induced by such an inner product should not be finer (nor weaker) than $\Vert~ \Vert_{\infty}$-topology, since both are going to be Banach-norm topology on $C[a, b]$. i.e. the new norm should be non comparable to the sup norm.

What would be such an inner product on $C[a, b]$ ? Any remark/suggestion is highly appreciated. Thanks.

  • $\begingroup$ Stop thinking of it as the space of continuous functions on $[a,b]$, and start thinking of it as $\Bbb R^{\aleph_0}$ in disguise. You are not required to support any existing structures on $C[a,b]$ other than the addition and scalar multiplication. The only thing the fact that the functions are continuous is useful for in this is that the values of the functions on a countable set are enough to completely determine the function, which means it is a vector space of countable dimension. $\endgroup$ May 8, 2021 at 13:37
  • $\begingroup$ @PaulSinclair $\mathbb R^{\mathbb N}$ is NOT countably dimensional! It is $c$-dimensional where $c$ is the cardinality of $\mathbb R$. $\endgroup$
    – Jochen
    May 8, 2021 at 14:34
  • $\begingroup$ @Jochen - $\Bbb N$ is countable. It is the very definition of "countable". And the actual expression I used, $\aleph_0$, is by definition the countably infinite cardinal. $\endgroup$ May 8, 2021 at 14:37
  • $\begingroup$ @PaulSinclair Come on! Of course, $\mathbb N$ is countable, but the usual definition of dimension refers to the cardinaltiy of Hamel bases where only finite linear combinations of basis vectors are allowed to represent all vectors. Then $\mathbb R^\mathbb N$ does not have a countable basis (and it is a bit tricky to show that the dimension is indeed $c$ without using the continuum hypothesis). $\endgroup$
    – Jochen
    May 8, 2021 at 14:50
  • $\begingroup$ But this problem is not about Hamel bases, is it? Instead, the very point is to produce an inner product on the space under which it is complete. That puts us in the perview of Schauder bases, not Hamel. $\Bbb R^{\aleph_0}$ clearly admits such a basis, which is the point of my comment. $\endgroup$ May 8, 2021 at 15:04

1 Answer 1


In principle, Paul Sinclair's first comment settles the question -- except for the dimension issues. The point is, that all separable Banach spaces are isomorphic as vector spaces because they have the same dimension $c$ (cardinality of $\mathbb R$). There is thus a linear bijective map $T:C[a,b]\to \ell^2$, the Hilbert space of square summable sequences. For two functions $f,g\in C[a,b]$ we can thus define a scalar product $\langle f,g\rangle_{C[a,b]}=\langle T(f),T(g)\rangle_{\ell^2}$. This makes $C[a,b]$ a Hlbert space: For a Cauchy sequence $f_n$ the sequence $T(f_n)$ is Cauchy in $\ell^2$ and hence converges to some $x$ in $\ell^2$ so that $f_n$ converges to $T^{-1}(x)$.

However, this scalar product has nothing to do with the usual topology of $C[a,b]$ of uniform convergence (e.g., because of the closed graph theorem). The existence requires the axiom of choice (to have Hamel bases), and for all concrete questions this existence result is completely useless.


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