I am studying certain quadratic forms on $L^0(m)$ equipped with the topology of (local) convergence in measure which in general is not locally convex. I am also interested in the situation where $m$ is not $\sigma$-finite and so $L^0(m)$ is not metrizable. It would be very nice to have a variant of the following elementary observation for the space $L^0(m)$.
When $(V,\tau)$ is a locally convex vector space the convergence of a net implies the convergence of suitable convex combinations. Namely, the following holds.
Proposition 1: Let $(V,\tau)$ be a locally convex vector space and let $(u_i)_{i\in I}$ be a net in $V$ with limit $u$. Furthermore, for each $i,j \in I$ let $ 0 \leq \lambda_{i,j} \leq 1$ be given such that
- for each $i$ we have $\lambda_{i,j} = 0$ for all but finitely many $j$,
- $\sum_{ j \geq i} \lambda_{i,j} = 1.$
Then the net
$$\left(\sum_{ j \geq i} \lambda_{i,j} u_j\right)_{i \in I}$$
converges to $u$.
Of course, the previous proposition in this strong form does not hold in general topological vector spaces. Nevertheless, it may hold for suitable subnets.
Question: Does the convergence statment of Proposition 1 hold in general topological vector spaces (in $L^0(m)$) for suitable subnets of $(u_i)$?
If $(V,\tau)$ is a metrizable topological vector space I do have the following result. To keep notation somewhat simpler I state it for Cesaro means.
Lemma 2: Let $(V,\tau)$ be a metrizable topological vector space and let $(u_n)$ be a convergent sequence with limit $u$. Then there exists a subsequence $(u_{n_k})$ such that for all of its subsequences $(u_{n_{k_l}})$ we have
$$\lim_{N \to \infty} \frac{1}{N} \sum_{l = 1}^N u_{n_{k_l}} = u.$$
Proof: Without loss of generality we may assume $u=0$. We choose a translation invariant metric $d$ with balanced balls around $0$ which induces $\tau$. Furthermore, we choose a subsequence $(u_{n_k})$ such that
$$\sum_{k = 1}^\infty d(u_{n_k},0) < \infty.$$
Now let $(u_{n_{k_l}})$ be an arbitrary subsequence of $(u_{n_k})$. The translation invariance of $d$ then implies
$$d\left(\frac{1}{N} \sum_{l = 1}^N u_{n_{k_l}}, 0\right) \leq d\left(\frac{1}{N} \sum_{l = 1}^M u_{n_{k_l}},0\right) + \sum_{l = M + 1} ^N d\left(\frac{1}{N}u_{n_{k_l}},0\right).$$
From this inequality and the balancedness of $d$-balls around $0$ we infer
$$d\left(\frac{1}{N} \sum_{l = 1}^N u_{n_{k_l}}, 0\right) \leq d\left(\frac{1}{N} \sum_{l = 1}^M u_{n_{k_l}},0\right) + \sum_{l = M + 1} ^\infty d\left(u_{n_{k_l}},0\right).$$
Choosing $M$ large enough and then using the continuity of the multiplication with scalars at $0$ finishes the proof.
