Let $\{f_n \}_n$ real valued functions such that: \begin{cases} f_n \geq 0 \\ \int_{\mathbb{R}} f_n =1\\ \forall \delta >0: \lim_{n \rightarrow +\infty } \int_{|t|>\delta}f_n(t)dt = 0 \end{cases} Show that, for every $p>1$: $$\lim_{n \rightarrow + \infty} ||f_n||_{L^p} = +\infty \quad \star$$
First I note that $1 = \int_{\mathbb{R}} f_n(t)dt = \int_{|t|<\delta}f_n(t)dt + \int_{|t|>\delta} f_n(t)dt$ and taking the limit follows that $$\lim_{n \rightarrow +\infty} \int_{|t|<\delta}f_n(t)dt=1$$
Now, in order to prove $\star$, I would like to show that for every $M>0$, there exists a $\bar{n} \in \mathbb{N}$ such that for every $n>\bar{n}$: $||f_n||_{L^p}^p>M^p$.
Therefore, since $f_n \geq 0$:
$$\int_\mathbb{R} f_n^p = \int_{|t|<\delta} f_n^p + \int_{|t|>\delta} f_n^p$$
Now I choose $\delta$ such that for $t \in \{t: |t|>\delta \}$ I have $f(t)<1$, therefore the second integral goes to $0$ as $n \rightarrow \infty$.
Also, since for $t \in \{t: |t|<\delta \}$ it holds that $f_n(t)>1$, and $f_n \geq 0$, it follows that $f_n^p>f_n$ for evry $p>1$.
So I can write $\int_{|t|<\delta}f_n^p > \int_{|t|<\delta}f_n$ and since the last term is $1$ as $n \rightarrow + \infty$, I have that $$\lim_{n \rightarrow \infty} \int_{|t|<\delta}f_n^p = 1 + \varepsilon_p$$ with $\varepsilon_p >0$
I don't know if it's okay up to now, but in any case I don't know how to conclude. Can anyone help me?
By Holder's inequality. $\int_{|t|<\delta} f_n(x)dx \leq \int (f_n^{p}(x)dx)^{1/p} (2\delta)^{1/q}$ where $\frac 1 p +\frac 1 q=1$. The conclusion follows easily from this since $\delta$ is arbitrary.