I have an exercise that says:
Let $BC(\Bbb R )$ the set of bounded and continuous functions from $\Bbb R $ to $\mathbb{F}$, where $\mathbb{F}$ is $\Bbb R $ or $\Bbb C $, endorsed with the following inner product $$ \langle f,g \rangle:=\sum_{m\geqslant 1}\frac{f(q_m)\overline{g(q_m)}}{2^m}\tag1 $$ where $(q_m)$ is an enumeration of $\Bbb Q $. Prove or disprove that $BC(\Bbb R )$ is a Hilbert space with this inner product.
I want to check if my counterexample below is correct and maybe if there is a more simple or straightforward counterexample.
I used the following theorem to construct my counterexample:
Theorem: let $(x_k)$ some sequence in a normed vector space $V$. Then $V$ is a Banach space if and only if $\sum_{k\geqslant 0}\|x_k\|<\infty\Rightarrow \sum_{k\geqslant 0}x_k$ exists in $V$.
Then assuming that $$ \sum_{k\geqslant 1}\|g_k\|=\sum_{k\geqslant 0}\sqrt{\sum_{m\geqslant 1}\frac{|g_k(q_m)|^2}{2^m}}<\infty\tag2 $$ we want to show that $\sum_{k\geqslant 1}g_k$ doesn't belong to $BC(\Bbb R)$.
Let a sequence $(g_k)$ in $BC(\Bbb R )$ defined by $g_k(x)=0$ if $x\notin [k,k+1)$ and $\|g_k\|_\infty =g_k(k+1/2)=k$ and let $(p_k)$ the increasing sequence of primes (that is, $p_0=2,\, p_1=3,\,p_2=5$, and so on) and define $N_k:=\{p_k^n: n\in \Bbb N_{>0}\}$. Then $(N_k)$ is a disjoint sequence of infinite subsets of $\Bbb N $ and because $p_k> k$ for all $k\in \Bbb N_{\geqslant 0} $ then we find that $\sum_{x\in N_k}2^{-x}< 2^{-k}$.
Then there is an injection $b:\Bbb Q \to \Bbb N$ such that the image of $b$ restricted to the set $[k,k+1)\cap \Bbb Q $ is $N_k$. Then by construction we find that $$ \sum_{q\in \Bbb Q }\frac{|g_k(q)|^2}{2^{b(q)}}\leqslant\frac{k^2}{2^k}\implies \sum_{k\geqslant 1}\|g_k\|\leqslant \sum_{k\geqslant 1}\frac{k}{2^{k/2}}< \infty\tag3 $$ However $g:=\sum_{k\geqslant 1}g_k$ is unbounded because $g(k+1/2)=k$ for all $k\in \Bbb N $, then by the theorem stated above $BC(\Bbb R )$ cannot be a Banach space, neither a Hilbert space.
UPDATE:
As pointed by @Daniel my counterexample is not right because it must work for any enumeration of the rationals and not just for a chosen enumeration.
Then I want to show that, for any enumeration of $\Bbb Q $, we can build a sequence $(f_k)$ with a similar behavior of the sequence $(g_k)$ of above. Let an enumeration $(q_n)_n$ of $\Bbb Q $ and set $Q_n:=\max\{q_1,\ldots ,q_n\}$ and define recursively the sequence of open intervals $$ I_1:=(q_1,q_1+1)\\ I_n:=(\max\{\sup I_{n-1},Q_n\},\max\{\sup I_{n-1},Q_n\}+1)\tag4 $$ Then by construction $(I_n)$ is a sequence of disjoint intervals of length one with the property that $\{q_1,\ldots ,q_n\}\cap I_n=\emptyset $. Now we set a sequence $(f_k)$ on $BC(\Bbb R )$ by $f_k(x)=0$ when $x\notin I_k$ and $\|f_k\|_\infty =k$. Then, by construction, each function $f_k$ have disjoint support and $$ \sum_{n\geqslant 1 }\frac{|f_k(q_n)|^2}{2^n}=\sum_{n\geqslant k+1}\frac{|f_k(q_n)|^2}{2^n}\leqslant \sum_{n\geqslant k+1}\frac{k^2}{2^n}=\frac{k^2}{2^k}\\ \therefore \quad \sum_{k\geqslant 1}\|f_k\|\leqslant \sum_{k\geqslant 1}\frac{k}{2^{k/2}}<\infty \tag5 $$ However by construction we have that for each $k\in \Bbb N $ there is some $x\in I_k$ such that $f_k(x)=k$, hence $\sum_{k\geqslant 1}f_k$ is unbounded, and by the theorem stated above it shows that $BC(\Bbb R )$ cannot be a Hilbert space.
Your modified construction works.
Here is another way to prove that $BC(\mathbb{R})$ is not complete with respect to $\lVert\,\cdot\,\rVert$:
We know that $BC(\mathbb{R})$ is a Banach space with respect to the supremum norm $\lVert\,\cdot\,\rVert_{\infty}$. Now for $g \in BC(\mathbb{R})$ we have $$\lVert g\rVert = \sqrt{\sum_{m = 1}^{\infty} 2^{-m}\lvert g(q_m)\rvert^2} \leqslant \sqrt{\sum_{m = 1}^{\infty} 2^{-m}\lVert g\rVert_{\infty}^2} = \lVert g\rVert_{\infty}\,.$$ Thus the identity is a continuous linear bijection $\bigl(BC(\mathbb{R}),\lVert\,\cdot\,\rVert_{\infty}\bigr) \to \bigl(BC(\mathbb{R}),\lVert\,\cdot\,\rVert\bigr)$, and the result follows by the open mapping theorem as soon as we have shown that the norms aren't equivalent.
There are many ways to do that. The easiest way I see is to consider functions $f_n\in BC(\mathbb{R})$ with $\lVert f_n\rVert_{\infty} = 1$ but $f_n(q_m) = 0$ for $m \leqslant n$, whence $\lVert f_n\rVert \leqslant 2^{-n/2}$.