Consider the Hilbert space of Dirichlet series $$\mathcal{H}^2=\lbrace f(s)=\sum_{n=1}^{\infty}\dfrac{a_n}{n^s}:\sum_{n=1}^{\infty}|a_n|^2<\infty\rbrace,$$ with norm $\|f\|_2^2=\sum_{n=1}^{\infty}|a_n|^2$. Why the Hilbert space $\mathcal{H}^2$ is not an algebra. That is, there exist an element $f\in\mathcal{H}^2$ and an element $g\in\mathcal{H}^2$ such that $fg\notin\mathcal{H}^2$.
If $f,g\in\mathcal{H}^2$, the Dirichlet series of $f$ has coefficients $a=\{a_n\}$ and the Dirichlet series of $g$ has coefficients $b=\{b_n\}$, then their product can be expressed by a Dirichlet series, with coefficients $c=\{c_n\}$, where $c=a*b$, the Dirichlet convolution of the sequences $a$ and $b$. In other words, $$c_n=a_n*b_n=\sum_{ij=n}a_i b_j.$$ By definition $a\in\ell_2$ and $b\in\ell_2$, where $\ell_2$ is the space of square-summable sequences. Is always $c\in\ell_2$? If not can you provide a counterexample?
I don't see an explicit example of $f, g\in \mathcal{H}^2$ such that $f\cdot g \notin \mathcal{H}^2$, but we can prove that such $f,g$ must exist.
Suppose to the contrary that for all $f,g \in \mathcal{H}^2$ their product $f\cdot g$ also belongs to $\mathcal{H}^2$. Then the product is a (commutative) bilinear map $\mu \colon\mathcal{H}^2 \times \mathcal{H}^2 \to \mathcal{H}^2$. Since convergence with respect to $\lVert\,\cdot\,\rVert_2$ implies pointwise convergence of the coefficients and $\mathcal{H}^2$ is a Banach space, the closed graph theorem asserts that $\mu$ is separately continuous. For, if for a fixed $g\in \mathcal{H}^2$ we have $\lVert f_k - f\rVert_2 \to 0$ and $\lVert f_k\cdot g - h\rVert_2 \to 0$, it follows that $h = f\cdot g$, since the coefficients of $f_k\cdot g$ converge to those of $f\cdot g$. But a separately continuous bilinear map on a product of Banach spaces is continuous by the Banach-Steinhaus theorem, so there is a $C\in (0,+\infty)$ with
$$\lVert f\cdot g\rVert_2 \leqslant C\cdot \lVert f\rVert_2\,\lVert g\rVert_2\,.\tag{$\ast$}$$
However, no such $C$ exists: For $\varepsilon > 0$, let
$$f_{\varepsilon}(s) = \sum_{k = 0}^{\infty} \frac{1}{2^{k(\varepsilon+s)}}\,,$$
so the coefficients are
$$a_n = \begin{cases} n^{-\varepsilon} &\text{if } n = 2^k\,, \\\; 0 &\text{if } n \text{ is not a power of } 2\,.\end{cases}$$
Then the coefficients $b_n$ of $f_{\varepsilon}\cdot f_{\varepsilon}$ are $0$ if $n$ is not a power of $2$, and
$$\sum_{m = 0}^k 2^{-m\varepsilon}\cdot 2^{-(k-m)\varepsilon} = (k+1)2^{-k\varepsilon}$$
for $n = 2^k$. We compute
$$\lVert f_{\varepsilon}\rVert_2^2 = \sum_{k = 0}^{\infty} 2^{-2k\varepsilon} = \frac{1}{1 - 2^{-2\varepsilon}}$$
and
\begin{align} \lVert f_{\varepsilon}^2\rVert_2^2 &= \sum_{k = 0}^{\infty} (k+1)^2\cdot 2^{-2k\varepsilon} \\ &= \sum_{k = 0}^{\infty} (k+2)(k+1) 2^{-2k\varepsilon} - \sum_{k = 0}^{\infty} (k+1)2^{-2k\varepsilon} \\ &= \frac{2}{(1 - 2^{-2\varepsilon})^3} - \frac{1}{(1 - 2^{-2\varepsilon})^2}\,, \end{align}
so
$$\frac{\lVert f_{\varepsilon}^2\rVert_2^2}{\lVert f_{\varepsilon}\rVert_2^4} = \frac{2}{1-2^{-2\varepsilon}} - 1$$
tends to $+\infty$ as $\varepsilon \to 0$, contradicting $(\ast)$.