PROBLEM
Proof if mixed strategy $\sigma^{*}\in \sum$ is s.t
$\forall i\in N : \operatorname{supp}(\sigma^{*}_{i})\subset B_{i}(\sigma^{*}_{-i}) \Rightarrow \sigma^{*}$ is Nash Equilibrium.
Notation:
_$G=(N,\{S_{i}\}_{i\in N},\{\prod_{i}\}_{i\in N})$
_$\prod_{i}$ Player's utility i; $\sigma_{i}\in \sum_{i}$ (Mixed Strategy); $s_{i}\in \S_{i}$ (Pure Strategy)
_ $supp(\sigma_{i})=\{s_{i}\in S_{i}: \sigma_{i}(s_{i})>0\}$
_ Best response correspondence.
$B_{i}:S_{-i}\to S_{i}$ s.t $B_{i}(s_{-i})=\{s^{*}_{i}\in S_{i}/\prod_{i}(s^{*}_{i},s_{-i})\geq\prod_{i}(s_{i},s_{-i})\forall s_{i}\in S_{i}\}$
I am trying to write the proof
Suppose that $\sigma^{*}$ t's not nash equilibrium
$\exists i \in N /\exists \sigma_{i}^{'}\in \sum_{i}:\prod_{i}(\sigma_{i}^{*},\sigma^{*}_{-i})<\prod_{i}(\sigma_{i}^{'},\sigma^{*}_{-i})$ ....$(\alpha)$
I know $\prod_{i}(\sigma_{i},s_{-i})=\sum_{s_{i}\in S_{i}}\sigma_{i}(s_{i}).\prod_{i}(s_{i},s_{-i})$ and in $(\alpha)$
$\sum_{s_{i}\in S_{i}}\sigma_{i}(s^{*}_{i}).\prod_{i}(s_{i},s_{-i})<\sum_{s_{i}\in S_{i}}\sigma^{'}_{i}(s_{i}).\prod_{i}(s_{i},s_{-i})$
Is the above correct?
but I do not know how to finish =(
It's sometimes helpful for these type of problems to not worry so much about notation at first, and formalize at the end.
Here, the answer is just going to come from the fact that, by definition, $i$ must be indifferent across all actions (i.e. $s_i$) in $B_i(\sigma_{-i})$, for all $\sigma_{-i}$, equilibrium or otherwise (and by the standard arguments there must be some $s_i \in B_i(\sigma_{-i})$ for all $\sigma_{-i}$). Thus by linearity of payoffs (von Neumann-Morgenstern utilities), the agent $i$ (for fixed $\sigma_{-i}$) must be indifferent between any pure strategy in $B_i(\sigma_{-i})$ and any mixed strategy whose support is contained wholly in $B_i(\sigma_{-i})$, and hence such mixed strategies must also lie in $B_i(\sigma_{-i})$, and thus your condition on supports reduces simply to saying that everyone plays a best response given what everyone else is doing, which is the definition of Nash equilibrium.
Another way of arguing this is geometrically: the best response correspondence (by linearity of payoffs) is going to be convex-valued. In particular, any convex combination of pure strategies in the best response (i.e. mixed strategies) must be in the best response too, and conversely if a mixture is a best response, it may be written as the convex combination of pure strategies in the best response.
Put formally:
$$ \forall i\in N \; \; \operatorname{supp}(\sigma^{*}_{i})\subset B_{i}(\sigma^{*}_{-i}) \iff \forall i\in N \; \; \sigma_i^* \in B_i(\sigma_{-i}^*) \iff \sigma^* \in \textrm{NE}(G) $$ where the first $\iff$ comes from the linearity of payoffs and the second $\iff$ from the definition of Nash equilibrium.