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Sorry to disturb. I have a question need some explanations from the experts on the MO-website.

As usual, we let $L(f,s)$ be the corresponding $L$-function associated to the newform $f$ on $SL_2(\mathbb{Z})$, of level $q$, $q\in \mathbb{N}^+$, say. It is known that, regarding the $s$-aspect subconvex bound, one has $L(f,1/2+it)\ll q^A (|t|+1)^{1/3+\varepsilon}$ for any $t \in \mathbb{R}$; see, e.g, A. Good's paper. Here, based on the functional equation, one sees that $A$ can be taken as $A=1/4+\varepsilon $; this corresponds to the convexity bound for $L(f,s)$ in the critical strip.

My question if we know a subconvex bound in $q$-aspect, whereby one can achieve a hybrid bound simultaneously in the s and q aspects? That is, whether or not the exponent $A$ here can be taken as an exponent which is less than 1/4?

If any expert knows some relevant knowledge upon this, please show some guides. Thanks in advance.

Many many thanks.

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    $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Commented May 17 at 7:49

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There are several confusions in your post.

1. Automorphic forms for the group $\mathrm{SL}_2(\mathbb{Z})$ have level $q=1$ by definition. You probably wanted to talk about newforms for the Hecke congruence subgroup $\Gamma_0(q)$.

2. Good only treated the case of $q=1$. The bound you claim was only proved recently (with an unspecified $A>0$) by Booker-Milinovich-Ng (2009) and Aggarwal (2020).

3. It is not known that $A=1/4+\varepsilon$ is admissible in the bound you claim. One can take $A=1/4+\varepsilon$ if at the same time the exponent of $|t|+1$ is enlarged to $1/2+\varepsilon$; this is the convexity bound.

4. As far as I know, that best hybrid subconvexity bound for the family at hand is due to Wu (2022): $$L(f,1/2+it)\ll_\varepsilon (q(|t|+1)^2)^{1/4-1/224+\varepsilon}.$$ If the nebentypus of $f$ is fixed, then $1/224$ can be improved to $25/1536$.

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  • $\begingroup$ Thanks for clarification of my these blurs. Much obliged. Could you give some more comments on how to understand "...if at the same time the exponent of ๐‘ž is enlarged to 1/2+๐œ€" in you post? Thanks. $\endgroup$
    – user528074
    Commented May 17 at 7:40
  • $\begingroup$ @user528074 Sorry, that was a typo. I meant that the exponent of $|t|+1$ is enlarged to $1/2+\varepsilon$. That is, the convexity bound is $L(f,1/2+it)\ll_\varepsilon(q(|t|+1)^2)^{1/4+\varepsilon}$. $\endgroup$
    – GH from MO
    Commented May 17 at 7:47
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    $\begingroup$ Thanks for timely reply. Great thanks. It is really illuminating to communicate with you. $\endgroup$
    – user528074
    Commented May 17 at 8:02

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