Show that the set of $f\in L^{1}$ so that $f\notin L^{p}$ is residual.

Question

Consider $L^{p}([0,1])$ with Lebesgue measure. Note that if $f\in L^{p}$ with $p>1$ then $f\in L^{1}$. Show that the set of $f\in L^{1}$ so that $f\notin L^{p}$ is residual.

First of all, a set $E\subset X$ is define to be generic if its complement is of the first category.

If we consider the set $$A_{N}:=\left\{f\in L^{1}: \displaystyle\int_{I}{|f|}\leq Nm(I)^{1-\frac{1}{p}}\mbox{ for all intervals $I$ }\right\}$$

How can I prove this set is nowhere dense?

Answer 1

Hint: If $\int |f|^p > N$, show there is $h \in L^\infty$ such that $\int h f > N$. Conclude that $\{f \in L^1:\; \int |f|^p > N\}$ is open. To show it is dense, consider adding to any $f \in L^1$ some function with small $L^1$ norm but large $L^p$ norm.

Answer 2

Proof of the fact that $A_N$ is nowhere dense: Fatou's Lemma shows that $A_N$ is closed in $L^{1}$; suppose $A_N$ has an interior point $g$. Then there exists $r>0$ such that $||f-g||_1 <r$ implies $f \in A_N$. By triangle inequality we get $\int_I |f| \leq 2N m(I)^{1-\frac 1 p} $ for any interval $I$ whenever $||f||_1 <r$. Take $I=(0,c)$ and $f(x)=ax^{\alpha}$ where $\alpha \in (-1,-\frac 1 p)$ and $0<a<r(1+\alpha)$. Then $||f||_1 <r$, and $\int_I |f| =\int_0^{c} ax^{\alpha} =a \frac {c^{1+\alpha}} {\alpha +1}$ so $a \frac {c^{1+\alpha}} {\alpha +1} \leq 2N c^{1-\frac 1 p} $. Dividing by $c^{1+\alpha}$ and letting $c \to 0$ we get a contradiction.