A textbook I own has the following question in it:
I'm having a very hard time finding a result for a), here's what I've done so far:
As the question has requested, I used the Cauchy-Schwarz inequality for integrals and deduced that |$\int$$_{B(x,r)}$u$^2$(x)dx|≤$\int$$_{B(x,r)}$|u(x)|$^2$dx<∞, furthermore by the mean value property our u(x)=$\frac{1}{α(n)r^n}$$\int$$_{B(x,r)}$u(y)dy hence I feel like this inequality can be shown by showing our $\int$$_{B(x,r)}$u(y)dy≤$\frac{C}{(α(n)r^n)^{1/2}}$ however I'm unsure how I could possibly find this result especially it's denominator.
Am I on the right track? I feel like I've hit a mental wall with this question and I'm unsure how I could go about completing it if I am? If not where is a better place to start?
The trick is finding the right functions to apply Cauchy-Schwarz to, namely $u$ and the indicator function of the ball. Using the mean value property and Cauchy-Schwarz, we get
$$\begin{align}|u(x)| &\leq \frac{1}{\alpha(n)r^n} \int_{B(x,r)} |u(x)|dx \\ &\leq \frac{1}{\alpha(n)r^n}\left(\int_{B(x,r)}dx\right)^{1/2}\left(\int_{\mathbb{R}^n} |u(x)|^2 dx\right)^{1/2} \\ &= \frac{C\cdot m(B(x,r))^{1/2}}{\alpha(n)r^n} \\ &= \frac{C}{\sqrt{\alpha(n)r^n}}\end{align}$$
Where $m$ is the Lebesgue measure.