I want to prove that $L^2$ is of the first category in $L^1$, thus I have to prove that $L^2$ is the countable union of nowhere dense subsets. The hint I get is: Take $g_n(x)=n$ for $0\leq x\leq\frac{1}{n^3}$ and show $\int{fg_n}\rightarrow 0$ for all $f\in L^2$ but not for every $f\in L^1$.
My question: What does this hint help for proving that $L^2$ is of the first category? I don't know how to use this hint. Can someone help me? :)
Thanks
Let $$A_{k}=\left\{f\in L^1 : \left|\int f g_k\right| \le 1 \right\}$$ and $$B_n = \bigcap_{k\ge n} A_k$$ Each set $B_n$ is closed in $L^1$. According to the hint, $$L^2\subset \bigcup_{n}B_n $$ It remains to show that each $B_n$ has empty interior. Given $f\in B_n$ and $\epsilon>0$, construct a function $h$ such that $\|h\|_{L^1}<\epsilon$ and $\int h g_k>2$ for some $k\ge n$. Observe that $f+g\notin B_n$.