Weak convergence in $L^q$ but not in $L^p$, where $1\leq pI've been trying to find a sequence $\{f_n\}$ of functions on $[0,1]$ that converges weakly in $L^q$ but does not converge weakly in $L^p$, where $1\leq p<q<\infty$. I'm stuck and any hints would be greatly appreciated.
(I find this to be tricky because the sequences are defined on $[0,1]$ and the functions cannot spread to infinity. Since we're working on $[0,1]$, we have $L^q\subset L^p$.)
The functions of the example below are not supported in $[0,1]$.
 A: Sorry for overlooking the requirement that the functions are supported on $[0,1]$. If this condition is added, then the statement is true.
Straight from the definition, $f_n$ converges weakly to $f\in L^q$ iff for every $g\in L^{q^*}$,
$$\int f_ng\to \int fg,$$
where $1/q+1/q^*=1$. Since $p<q$, $p^*>q^*$, so $L^{p^*}([0,1])\subset L^{q^*}([0,1])$.
Then the above convergence holds for every $g\in L^{p^*}$. This implies $f_n$ converges weakly to $f$ in $L^p$. As a sanity check, we see that $f$ is in $L^p$ because $L^q([0,1])\subset L^p([0,1])$.
More succinctly, this is because continuous linear maps between Banach spaces are also weakly continuous (See Theorem 1.1 of Chapter 6 of Conway, A Course in Functional Analysis).
Just for reference, the result is false if the functions are supported on $\mathbb R$.
Let $r\in(p,q)$ and $f_n=n^{-1/r}$ on $[0,n]$ and $f_n=0$ elsewhere. Then
$$\|f_n\|_{q}=(n\cdot n^{-q/r})^{1/q}=n^{1/q-1/r}\to 0$$
as $n\to\infty$, so $f_n\to0$ in $L^q$. On the other hand, we replace $q$ by $p$ in the above inequality to get
$$\|f_n\|_{p}=n^{1/p-1/r}\to\infty$$
as $n\to\infty$. By the uniform boundedness principle (among other approaches), $f_n$ does not converge weakly in $L^p$.
