Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of functions in $L^1(\mathbb{R})$ such that:
- $f_n \geq 0$;
- $\int f_n = 1$;
- for all $\delta >0$, $\lim_{n\to+\infty}\int_{|t|\geq \delta} f_n(t)\mathrm{d}t =0$.
I want to show that $\lim _{n\to+\infty}\|f_n\|_{L^p}=+\infty$, for all $p>1$.
My idea was to show that for each $n$ there is a sufficiently large subset of $[-\delta,\delta]$ in which the function $f_n$ is very large. To be more precise, fixed $\delta>0$ and $m\in\mathbb{N}$, there exists $N_{\delta,m}$ such that $$ \int_{|t|\geq \delta} f_n(t)\mathrm{d}t < \frac{1}{m}, \quad n\geq N_{\delta,m}. $$ Since $\int f_n = 1$, the set $ A_{n,m,\delta} = \{ t\in[-\delta,\delta], \; f_n(t)>\frac{2}{\delta}\left(1-\frac{1}{m}\right) \} $ must have strictly positive measure, otherwise $\int f_n < 1$. So, $$ \|f_n\|_{L^p}\geq \int_{A_{n,m,\delta}}|f_n(t)|^p \mathrm{d}t\geq |A_{n,m,\delta}| \frac{2^p}{\delta^p}\left(1-\frac{1}{m}\right)^p ,$$ and therefore for sufficiently small $\delta$, $\|f_n\|_{L^p}$ would be large. However, $|A_{n,m,\delta}|$ is bounded below with zero, so I fear that this argument cannot work.
Does anyone know how to fix my argument or if there is a smarter way to prove that $\lim _{n\to+\infty}\|f_n\|_{L^p}=+\infty$?