Assume that you have a jointly meausrable stochastic process with $P(\int_o^T|X_s|ds<\infty=1)$. And for this stochastic process you are able to find a sequence of product measurable stochastic processes $\phi_n$, with $P(\int_o^T|\phi_{s,n}|^2ds<\infty=1)$. And $\int_o^T|\text{sgn}(X_s)\sqrt{|X_s|}-\phi_{s,n}|^2ds\rightarrow 0$ in probability.
Does it then follow that: $\int_0^T|X_s-\text{sgn}(\phi_{s,n})\phi_{s,n}^2|ds\rightarrow 0$ in probability?
It seems very natural that this should be the case, and I would imagine that it should follow pretty straight forward from Hölder's inequality or something like that. But I am not able to actuarlly finish the argument.
attempt:
First I try to use the triangle inequality on the integrand.:
$\int_o^T|X_s-\text{sgn}(\phi_{s,n})\phi_{s,n}^2|ds\le \int_0^T|\text{sgn}(X_s)\sqrt{|X_s|}\sqrt{|X_s|}-\sqrt{|X_s|}\phi_{s,n}|ds+\int_0^T|\sqrt{|X_s|}\phi_{s,n}-\text{sgn}(\phi_{s,n})\phi_{s,n}^2|ds=\int_0^T \sqrt{|X_s|}|\text{sgn}(X_s)-\phi_{s,n}|ds+\int_0^T|\phi_{s,n}||\sqrt{|X_s|}-\text{sgn}(\phi_{s,n})\phi_{s,n}|ds$.
Using Hölder on the first integral: $\int_0^T \sqrt{|X_s|}|\text{sgn}(X_s)-\phi_{s,n}|ds\le\int_0^T|X_s|ds^{0.5}\int_0^T|\text{sgn}(X_n)-\phi_n|^2ds$. Since $P(\int_o^T|X_s|ds<\infty=1)$, given $\epsilon_1,\epsilon_2$ there is an M, such that $P(\int_o^T|X_s|ds<M^{0.5})<\epsilon_2 /2$, there is an $N$ so that for $n \ge N$ $P(\int_0^T |\text{sgn}(X_s)\sqrt{|X_s|}-\phi_n|^2>\epsilon_1/M)<\epsilon_2/2$. Then for $n \ge N$ $P(\int_0^T \sqrt{|X_s|}|\text{sgn}(X_s)-\phi_{s,n}|ds>\epsilon_1)\le P(\int_0^T|X_s|ds^{0.5}\int_0^T|\text{sgn}(X_n)-\phi_n|^2ds>\epsilon_1)<\epsilon _2$. I am a little unsure but I think this shows that the first integral converges to 0 in probability.
But what about the second integral?: $\int_0^T|\phi_{s,n}||\sqrt{|X_s|}-\text{sgn}(\phi_{s,n})\phi_{s,n}|ds$, does this converge to 0 in probability?
May it be that what I wrote above does not hold?