Let $G$ be a split reductive group over an arbitrary field $k$. By definition, see Jantzen (*), an ascending chain $$0 = V_0 \subset V_1 \subset V_2 \subset \dots$$ of submodules of a $G$-module $V$ is called a Weyl filtration of $V$ if $V = \bigcup_{i \geq 0} V_i$ and if each $V_i/V_{i-1}$ is isomorphic to some Weyl module $V(\lambda_i)$ of highest weight $\lambda_i$.

It seems that if a module admits such a Weyl filtration, we have $V \cong \bigoplus_{i} V(\lambda_i)$. Indeed, according to remark II.4.19 in Jantzen, we can order the $\lambda_i$ such that $\lambda_i < \lambda_j$ implies $i > j$. The decomposition $V \cong \bigoplus_{i} V(\lambda_i)$ then follows by induction from $\operatorname{Ext}^1_G(V(\lambda),V)=0$ if $V$ has no weight $\mu > \lambda$ (see Jantzen II.2.14 Remark 2).

In particular I would like to use that we can write the symmetric square of a Weyl module as a direct sum of Weyl modules and that we can derive the terms from Weyl's character formula. This should follow from Proposition II.4.21 in Jantzen which basically says (the dual of) if $V$ admits a Weyl filtration then so does $V \otimes V$.

However, this results seems quite standard and I am quite surprised that Jantzen doesn't mention it (as far as I know). So my question is whether there is a more direct approach towards this problem and whether anyone knows a good reference.

EDIT. The reasoning in the second paragraph is not correct. We would want the $\lambda_i$ to be ordered such that if $\lambda_i < \lambda_j$ then $i < j$. I am not sure whether this is possible. Probably it is not, which explains why this isn't in Jantzen.

(*) Jantzen, Representations of Algebraic Groups (AMS, 2003)