@PeterHumphries Dear Humphries, you said if the newform $f$ is of square-free level $q$ with trivial nebentypus, and the form $g$ is of level $r^2$, $(r,q)=1$ with trivial nebentypus, such that there exists a primitive Dirichlet character $\Psi\bmod r$ such that $g\otimes \Psi$ has the level $r$, then this is doable. But, as I know, the level of $g\otimes \Psi$ should be $r^2 $ by H. Iwaniec's topics-book. It seems a little weird.

@PeterHumphries Okay, thanks for useful explanations.

@PeterHumphries Dear Humphries, could you please show some references on this, if $g$ is of level $r^2$ as you assumed? On the other hand, I also curious if Nelson's work can be adaptable to show that the level aspect subconvexity of $f$ for $L(\text{sym}^2 f\otimes \chi)$ implies the aspect subconvexity of $f$ for $L(\text{sym}^2 f\otimes g)$, if $\chi$ and $g$ are all allowed to be taken as certain special cases. Great thanks!

Sorry, .....we see that $a_f(n)$... , $a_f(n)$ should be $a_f(n)=\lambda_f(n)n^{1/4}$.

Great answer! According to the normalization, we see that $a_f(n)=\lambda_f(n)n^{-1/4}$. in other words, one has the analog of square-root cancellation. That is, $$\sum_{n\le X}a_f(n)e(n)\alpha \ll X^{1/2}\log (2X).$$ However, could you please show me how $f(z)\ll y^{-(k+1/2)/2}$ if $f$ is a cusp form of half-weight $k+1/2$?? Could you show me certain reference? If $f$ is a cusp form of integral weight $k$, we know $f(z)\ll y^{-k/2}$ .

@Kimball Sorry, it is a bit overstated...

Great thanks for the great answer!

@OfirGorodetsky Thanks for comment.