$L^p$ convergence is equivalent for $p \geq 0$

Question

Is $L^p$ convergence of a sequence of functions $f_n$ equivalent for $p \geq 1$? In other words, can I say that $f_n$ converges in $L^1$ if and only if it converges in $L^2$? Otherwise, please provide a counterexample.

Answer 1

No: consider $[0,1]$ with Lebesgue measure, and define $f_n=n^21_{[0,\frac{1}{n^3}]}$. Then $$\int_0^1f_n(x)\;dx=n^2\cdot\frac{1}{n^3}=\frac{1}{n} $$ so $f_n\to 0$ in $L^1$, but $$\int_0^1f_n(x)^2\;dx=n^4\cdot\frac{1}{n^3}=n$$ for all $n$, so $\{f_n\}$ does not converge in $L^2$.

However, if $(X,\mathcal{M},\mu)$ is a probability space, then $$ ||f||_p\leq ||f||_q$$ if $0<p\leq q$, so convergence in $L^q$ implies convergence in $L^p$.

Answer 2

No. An easy way to see this is that, by the construction of $L^p$ in terms of step functions, continuous functions are dense in any $L^p$ space, but there are functions in the set $L^2 \setminus L^1$, for example $x^{-1}$ is in $L^2(1,\infty) \setminus L^1(1,\infty)$. Hence there is a sequence of continuous functions that converge to $x^{-1}$ in $L^2(1,\infty)$, but not in $L^1(1,\infty)$ (e.g. take the function that is $x^{-1}$ for $x<n$ and $0$ otherwise).