# Uniform integrability contradicts convergence to $L^2$ subspace

Let $$V$$ be a finite-dimensional subspace of $$L^2(\mathbb{R})$$.

Assume that $$f_n$$ is a sequence of square-integrable functions with $$\Vert f_n \Vert_{L^2}=1$$ that satisfies two properties:

1.) $$d(f_n,V) \rightarrow 0$$ that is the distance to $$V$$ vanishes in the limit

2.) There exists a uniform (in $$n$$) constant $$k$$ and a strictly positive function $$g$$ such that the following uniform integrability condition holds $$\int_{\mathbb{R}} g(x) \vert f_n(x) \vert^2 \ dx \le k.$$

I want to show that if for all $$v \neq 0$$ in $$V$$ we have

$$\int_{\mathbb{R}} g(x) \vert v(x) \vert^2 \ dx=\infty$$ then such a sequence $$f_n$$ cannot exist.

The intuition is that the $$f_n$$ are more and more supported in $$V$$ where every element has infinite integral against $$g$$, so the uniform integrability condition cannot hold.

EDIT: If we knew for example that $$f_n$$ would not just converge to $$V$$ but to a fixed element $$f$$ in $$V$$, then it would follow that for a subsequence of the $$f_n$$ we would have $$f_n \rightarrow f$$ almost everywhere and thus get a fast contradiction using Fatou's lemma.

The question was then deleted by the OP while I was typing the answer. I thought the question might still be of some interest and will give an answer to it below.

Let $$g_n:=P_V f_n$$, where $$P_V$$ is the orthoprojector onto $$V$$. Then $$\|g_n\|_2\le1$$, $$g_n\in V$$, and $$V$$ is finite dimensional. So, passing to a subsequence, without loss of generality we may assume that $$g_n\to v$$ for some $$v\in V$$. Also, $$\|f_n-g_n\|_2=d(f_n,V)\to0$$. So, $$f_n\to v$$ in $$L^2$$ and hence $$\|v\|_2=1$$, so that $$v\ne0$$ and hence $$\int g|v|^2=\infty. \tag{1}$$ Also, in view of the condition $$f_n\to v$$ in $$L^2$$, passing again to a subsequence, without loss of generality we may assume that $$f_n\to v$$ almost everywhere. Now the Fatou lemma and condition (1) imply $$\int g|f_n|^2\to\infty,$$ as desired.