I am stuck on the following problem:
Let $B_n$ be the unit ball in $\mathbb{R}^n$. Therefore, if $f \in L^2(B_n)$, then $|x|^{-1} f(x)\in L^1(B)$ for $n \geq 3$.
My work:
Define $g(x) = |x|^{-1} f(x)$. For $g \in L^1(B)$, we need $\int |g| d \mu < \infty$. Which is the same as $g$ being integrable.
By Hölder's inequality:
$$\int_{B_n} |g| d\mu = \int\left||x|^{-1}f(x) \right| d\mu \leq \left| \left| \frac{1}{|x|} \right| \right|_p ||f||_q$$
for $1/p + 1/q= 1$. We can pick, $p = 2$ since we know that $||f||_2 < \infty$. This means that $q=2$, and we have
$$\left| \left| \frac{1}{|x|} \right| \right|_2=\left| \int_{B_n}\frac{1}{|x|^{1/2}} d\mu \right|^2$$
We know that $\int_{B_n} f(x) d\mu < \infty$ when $|f(x)| \leq \frac{c}{|x|^p}$ on $B_n$ for some constant $c$ and $p<n$. This is clearly true for the integral above, since $1/2 < 1 \leq n$.
Now, I know I am doing something wrong because in no part of my argument do I require $n \geq 3$, but I don't know why I messed up.
You did use it: $\displaystyle\int_{B_{n}}\dfrac{1}{|x|^{\alpha}}dx<\infty$ if and only if $\alpha<n$. In this case, you chose the $\alpha$ to be $2$, which is smaller than the dimension $n=3$.