This question is out of Rudin Functional analysis, but only attention was really brought to part (b) on this site, my apologizes if this is considered a re-post. I was just hoping for some assistance. I added what I have so far in my proof.
Let $L^1$ and $L^2$ be the usual Lebegue space on the unit interval. Prove that $L^2$ is of the first catagory in $L^1$, by the following.
Show that $\{f:\int |f|^2 \leq n\}$ is closed in $L^1$ but has empty interior.
proof:
Let $f \in L^2$ be arbitrary. Then $\int |f|^2 d \mu \leq n$, for some finite $n$. Observe that $|f g| \leq \frac{|f^2 + g^2|}{2}$. Suppose that $g = 1$. Then $$ \int |fg| d\mu = \int |f| d \mu \leq \frac{1}{2}\int |f^2 + 1| d\mu \leq \frac{1}{2}\int |f|^2 d\mu + \frac{1}{2} \int d \mu \leq \frac{1}{2}(n + 1) $$ Hence $f \in L^1$ $f$ arbitary and closed in $L^2$ implies that $L^2$ is closed in $L^1$.
I'm really struggling with showing that the interior is empty, and not sure how to make the argument.
