Sequence of $L^2$ functions converging to zero weakly s.t. $|f_n|^2$ converges to 1 weak-star?

Question

I am trying to construct a sequence $\{f_n\} \in L^2([0,1])$ with $f_n \geq 0$ a.e. such that $f_n \to 0$ weakly in $L^2$ (meaning $\int f_n dx \to 0$ for all $f \in L^2([0,1]))$ and such that for all $g$ in $C([0,1])$, i.e. $g$ continuous, $$ \int |f_n|^2 g dx \to \int g dx. $$ My instinct was that such thing is possible but the more I look at it, the more I find obstacles. Such a sequence must be quite delicate in that if we strengthen the conclusion to $\int |f_n|^2 g dx \to \int g dx$ for all $g \in L^\infty$ then the claim is immediately false as Egorov's theorem gives that (as $f_n \to 0$ in measure implies $f_n^2 \to 0$ in measure and convergence in measure implies convergence a.e. along a subsequence) there exist sets $B$ of close arbitrarily close to full measure along which $\int f_n 1_B dx \to 0$.

So I suppose my question is: does such a sequence of functions exist? If so, can it be defined explicitly? (And if such a sequence cannot exist, I'd welcome a proof to that effect)

Edit: I intended to require that the functions are real-valued and $f_n \geq 0$ so in particular weak convergence does imply convergence to zero in measure.

Edit 2: I don't "get" MO apparently seeing as I've spent 20 mins trying to work out how to thank @AleksieHulikov for his answers and it seems I don't have the "reputation" needed to comment nor to accept his answer. Kids these days, sigh.

Anyway: I would like to thank Aleksie. For the record this was very much not FA homework (though it might make a good hw problem for those of you teaching that), it was simply that I haven't thought hard about analysis in decades (since grad school to be precise) and when I ran into this, some of my younger colleagues who I asked suggested I come here.

Answer 1

The answer to the updated question is still no, with a following (only slightly less trivial) counterexample: take $$f_n(x) = \sqrt{n}\sum_{k = 0}^{n-1} \chi_{[\frac{k}{n}, \frac{k}{n} + \frac{1}{n^2}]}(x)$$

We have $\|f_n\|_{L^2} = 1$ and $\|f_n\|_{L^1} = \frac{1}{\sqrt{n}}\to 0$, so $f_n$ weakly converges to $0$ in $L^2$. On the other hand, to prove that the integral of $f_n^2g$ converges to the integral of $g$ you can approximate $g$ in the uniform norm by the linear combination of the characteristic functions of the intervals, and for $g(x) = \chi_{[a, b]}(x)$ that the limit is equal to $b - a$ is easy: up to the error of order $\frac{1}{n}$ (coming from the intervals containing $a$ and $b$), we sum $[n(b-a)]$ terms each of which is equal to $\frac{1}{n}$, which is (again, up to the error of order $\frac{1}{n}$) gives us exactly $b-a$. Alternatively, you can use uniform continuity directly to get the same result.