If $\nu^{\ast 2}$ is a tight probability measure, is $\nu$ itself tight? Let $E$ be a normed $\mathbb R$-vector space and $\mu$ be a tight$^1$ probability measure on $\mathcal B(E)$. Assume $\nu$ is another probability measure on $\mathcal B(E)$ and$^2$ $$\mu=\nu^{\ast k}\tag1$$ for some $k\in\mathbb N$.

Are we able to show that $\nu$ is tight as well?

I know how we can prove that the convolution of tight measures is tight and how we can show that if $\nu_1$ and $\nu_1\ast\nu_2$ are tight, then $\nu_2$ is tight as well, but I'm not sure how we could prove the desired claim here.
(To give some context: I would like to show that if $\mu$ is infinitely divisible (i.e. for all $n\in\mathbb N$, there is a probability measure $\nu_n$ such that $\mu=\nu_n^{\ast n}$) and tight, then the convolution roots $\mu^{\ast\frac1n}$ are well-defined for all $n\in\mathbb N$. In order to show uniqueness, we need that the $\nu_n$ are tight.)

$^1$ i.e. for all $\varepsilon>0$, there is a compact $K\in\mathcal B(E)$ with $\mu(K^c)<\varepsilon$.
$^2$ If $\nu_1,\ldots,\nu_k$ are measures on $\mathcal B(E)$ and $$\theta_k:E^k\to E\;,\;\;\;x\mapsto x_1+\cdots+x_k,$$ then the convolution of $\nu_1,\ldots,\nu_k$ is defined to be the pushforward measure $$\nu_1\ast\cdots\ast\nu_k:=\theta_k(\nu_1\otimes\cdots\otimes\nu_k)$$ of the product measure $\nu_1\otimes\cdots\otimes\nu_k$ with respect to $\theta_k$. If $\nu_1=\cdots=\nu_k$, we simply write $\nu_1^{\ast k}:=\nu_1\ast\cdots\ast\nu_k$.
 A: As noted in the comments, there are some potential issues for non-separable spaces, since then it might happen that the product sigma algebra on $E \times E$ is a strict subset of the Borel sigma algebra on $E \times E$. In this case, it is not clear that the map $\theta : E \times E \to E, (x,y) \mapsto x+y$ is measurable (with respect to the product sigma algebra), so that one can indeed define the convolution $\mu \ast \nu$ as the pushforward of the product measure $\mu \otimes \nu$ under $\theta$.
In the following, I therefore ignore some measurability issues. In case these are satisfied/resolved, I prove the following statement: If $\mu,\nu$ are probability measures on $E$ such that $\mu \ast \nu$ is tight, then $\mu$ and $\nu$ are tight as well. By symmetry, it suffices to show that $\nu$ is tight.
Since $\mu \ast \nu$ is tight, given $\epsilon > 0$, there is a compact set $K \subset E$ such that
$$
1-\epsilon
< \mu \ast \nu (K)
= \int_E \int_E 1_K (x+y) \, d \nu(y) \, d \mu(x)
= \int_E \nu(K-x) \, d \mu(x).
$$
This easily implies that there exists $x \in E$ such that $\nu(K-x) > 1 - \epsilon$. Since $K-x$ is compact and $\nu$ is a probability measure (so that $\nu(E \setminus (K - x)) < \epsilon$), we are done.
A: As mentioned in the comments, we must assume $\mathrm{E}$ is separable or else the convolution may not even be well-defined.
Consider the probability space $(\mathrm{E}, \mathscr{B}_\mathrm{E}, \nu).$ On this probability space, the identity function $I_\mathrm{E}$ is an $\mathrm{E}$-valued random object whose distribution is $\nu.$ A fortiori, there exists random objects with law $\nu.$ Furthermore, $\nu^{*k}$ is, by definition, the law of $X_1 + \ldots + X_k$ where each the $(X_j)$ are jointly independent and each one of them has law $\nu.$ The function $\mathrm{E}^k \to \mathrm{E}$ given by $(e_1, \ldots, e_k) = e_1 + \ldots + e_k$ is continuous and as such, for a given compact set $\mathrm{K}$ of $\mathrm{E},$ the set $\mathrm{K} + \ldots + \mathrm{K}$ is also compact in $\mathrm{E}.$ Clearly,
$$
\bigcap\limits_{j = 1}^k \{X_j \in \mathrm{K} \} \subset \{X_1 + \ldots + X_k \in \mathrm{K} + \ldots + \mathrm{K}\}.
$$
As a consequence, by taking complements and using that the probability of a union is less or equal than the sum of the individual probabilities,
$$
\begin{align}
\nu^{*k}\left( (K + \ldots + K)^\complement \right) &= \mathbf{P}(X_1 + \ldots + X_k \notin \mathrm{K} + \ldots + \mathrm{K}) \\
&\leq \sum_{i = 1}^ k \mathbf{P}(X_i \notin \mathrm{K}) \\
&= k \nu \left( \mathrm{K}^\complement \right).
\end{align}
$$
Since $\nu$ is tight, for all $\varepsilon > 0$ there exists a compact set $\mathrm{K}$ such that $\nu(\mathrm{K}) \geq 1 - \dfrac{\varepsilon}{k}$ and so,
$$
\nu^{*k}(\mathrm{K} + \ldots + \mathrm{K}) = 1 - k \nu \left( \mathrm{K}^\complement \right) \geq 1 - \varepsilon.
$$
Thus, $\nu^{*k}$ is also tight. Q.E.D.
Scholium. As I mentioned in the comments, I do not think that a tight probability measure on an infinite-dimensional normed space is very interesting for it has to assign total mass to a set of the form $\bigcup\limits_{n \in \mathbf{N}} \mathrm{K}_n$ where each $\mathrm{K}_n$ is a compact subset, and as such, it must have empty interior (it has to be very flat and perhaps finite-dimensional).
