I'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 $[0,1]$.


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


as $n\to\infty$. By the uniform boundedness principle (among other approaches), $f_n$ does not converge weakly in $L^p$.

  • $\begingroup$ The example seems actually impossible: Let p*, q* be the conjugate exponents of p, q resp. Since [0,1] contains sets of measure at most 1, if p<q (i.e., p*>q*) then L^q([0,1]) is a subset of L^p([0,1]) and L^{p^*}([0,1]) is a subset of L^{q^*}([0,1]). Therefore, if f_n are in L^q, hence also in L^p, then weak convergence in L^q implies weak convergence in L^p. $\endgroup$ – Aubrey Mar 10 '15 at 2:33
  • $\begingroup$ Thank you! Perhaps you should just change the phrase "the statement is true" to "the statement is false". $\endgroup$ – Aubrey Mar 10 '15 at 2:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.