An estimate of the second moment of the Ramanujan $\tau$- function

Question

Let $\tau$ the Ramanujan $\tau$-function, it is proved here that $$\sum_{x\le p\le x+x^\theta}\tau(p)^2p^{-11}\log p\gg x^\theta\quad (1)$$ for some $\theta<1.$ My question is: how we can infer from that result the following $$\sum_{x\le p\le x+x^\theta}\tau(p)^2p^{-11}\gg x^{\theta}\quad (2)$$

Answer 1

As your paper explains p.3 $$\sum_{p < x} \tau(p)^2 p^{-11} \log p \sim x,\qquad \sum_{p < x} \tau(p)^2 p^{-11} \sim \frac{x}{\log x} $$ is the prime number theorem for this L-function

thus $\sum_{p \in [x,x+x^\theta]} \tau(p)^2 p^{-11}\gg x^\theta$ can't be true, since it would imply $$\sum_{p < x}\tau(p)^2 p^{-11} \gg \int_2^x \frac{ \sum_{p \in [y,y+y^\theta]} \tau(p)^2 p^{-11}}{y^\theta} dy \gg x$$