If the limit of a $L^2$ sequence is in $L^\infty$, is the sequence bounded in $L^\infty$?

Question

Let $f_n \to f$ in $L^2$ on a bounded domain. We know that $f \in L^\infty$.

Does it follow that $\lVert f_n \rVert \leq A$ for a constant $A$ independent of $n$, for a subsequence if necessary?

I think it is true since the functions get closer to the limit $f$, and at worst, $f_1$ is the furthest away. The sequence cannot oscillate due to the pointwise a.e. convergence for a subsequence.

Answer 1

Consider $f_n = n\chi_{[0,1/n^3]}$ on $[0,1].$

Answer 2

Take a smooth function which is supported on $[0, e^{-n}]$ and with height $n$. One way to do this is to take the characteristic function of $[e^{-n}/4, 3e^{-n}/4]$ and mollify it, then multiply by $n$. Then these functions converge to zero in not only $L^2$, but also every $L^p$ with $p < \infty$. They're each compactly supported, smooth, and bounded; yet still their $L^{\infty}$ norms blow up.

So there is no reasonable variation on the hypotheses to get the desired conclusion.