L1 and L2 norm not equivalent on sequence of functions where H^1 norm is uniformly bounded

Question

I am looking for a sequence of $f_n\in L^2(0,T)$ where

1. $\|f_n\|_{H^1(0,T)}\leq M \quad \forall n \in \mathbb{N}$
2. $(\forall C>0)(\exists n\in \mathbb{N})(\|f_n\|_{L^2(0,T)} > C\|f_n\|_{L^1(0,T)}) $

The first condition is weaker than $L^2$ convergence. The second condition reminds me of function sequences used to show $L^1(0,T)$ and $L^2(0,T)$ are not equivalent.

I thought I had succeeded with $f_n=\sqrt{n}(t/T)^n$. I believe this function satisfies 2. Also, $\lim_{n\to \infty} \|f_n\|^2_{L^2(0,T)} = T/2$. However, the $H^1$ norm blows up.

Answer 1

Take $$ f_n(t) = \max\left(0, \frac1n-t\right). $$ Then $|f_n(t)|, |f_n'(t)|\le 1$, the function $f_n$ is supported on $(0,1/n)$, and the $L^2$ and $L^1$ norms are $$ \|f_n\|_{L^1} = \frac1{2n^2}, \ \|f_n\|_{L^2} = \sqrt{ \frac1{3n^3}}. $$