Let $p \in (1, \infty).$ Then there is a sequence $f_k$ satisfying
1) $f_k \in L^q$ for all $1 \leq q < \infty$
2) $f_k$ converges to $0$ in $L^q$ for all $q \in [1,p)$
3) $f_k$ diverges in $L^p$
$\textbf{Attemp}$ Since the integral for $1 \leq q < p$ converges to $0$ and at exactly $p$ it diverges, I think of $\int \frac{1}{x^k}$ which can be either $\log(x)$ and $\frac{1}{x^l}$ where $\log(x)$ diverges at infinity and $\frac{1}{x^l}$ converges to $0$ as $x \rightarrow \infty.$
So I set $f_k(x) = \int_1^k g(x) dx$ such that $f_k(x) = \frac{1}{k^{l_k}}$ for some $l_k > 0$ for all $q < p$ and $f_k(x) = \log(k)$ when $q = p.$
Somehow I feel it is impossible to find such $g$ since $q < p.$
Any suggestion ? Any idea on maybe a better functions ?
You are one the right track. Take $f_k(x)=x^{-1/p}1_{[1/k^2,1/k]}$, where $1_{I}$ is the indicator function of the interval $I\subset \mathbb{R}$.
1) Is obviously satisfied.
2)$\Vert f_k\Vert_q^q=\int_{1/k^2}^{1/k}x^{-q/p}dx = \frac{1}{1-q/p}\left((1/k)^{1-q/p} - (1/k^2)^{1-q/p} \right) \rightarrow 0$ as $k\rightarrow \infty$, when $q < p$.
3) $\Vert f_k \Vert_p^p=\int_{1/k^2}^{1/k}x^{-1}dx=\ln(1/k)-\ln(1/k^2)=\ln(k)\rightarrow \infty$, as $k\rightarrow \infty$.