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.

F-norms(look at Köthe's or Jarchow's book) which induces the topology? Perhaps you can then use the idea of Lemma 2. $\endgroup$ – Jochen Wengenroth Sep 24 '15 at 20:56subnetis quite different from that of asubsequence. In particular, the index set of the subnet can be very different. I don't know however, if this is helpful. $\endgroup$ – Jochen Wengenroth Sep 25 '15 at 10:20