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Let $|t|\geqslant 3$. I don't know the bound for $L$-functions when $\chi$ is a primitive character modulo $q$. It seems to me that $$\zeta(\tfrac{1}{2}+it)\ll_{\varepsilon} |t|^{1/6+\varepsilon}$$ and similarly $$L(\tfrac{1}{2}+it,\chi)\ll_{\varepsilon} (q|t|)^{1/6+\varepsilon}\;\text{ and } \;\sum_{\chi(q)}\int_{|t|\leqslant T} |L(\frac{1}{2}+it,\chi)|^4dt\ll_{\varepsilon} (qT)^{1+\varepsilon}.$$

In Titschmarsh book, $\mu(\sigma)$ defined as an infimum of $\xi$ for $\zeta(\sigma+it)\ll |t|^{\xi}$, states that $\mu(\sigma)\leqslant \frac{1}{2}-\frac{\sigma}{2}$ if $\sigma\in [1/2,1-\varepsilon]$. I want to bound $$L(\sigma+it,\chi)\ll_{\varepsilon} (q|t|)^{1/6+\varepsilon}\;\text{ and } \;\sum_{\chi(q)}\int_{|t|\leqslant T} |L(\sigma+it,\chi)|^4dt\ll_{\varepsilon} (qT)^{1+\varepsilon}$$ for all $\sigma\geqslant 1/2$. Is it possible for these bounds for $\sigma>1/2$? Or how to obtain the bound in terms of $\sigma$? Shouldn't this holds for all $\sigma>1/2$ since it is greater than $1/2$?

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    note that $\mu$ is convex and decreasing in $\sigma$ - convexity is true for any holomorphic function that is polynomially bounded on vertical lines so $\mu$ makes sense, while decreasing is true when $f$ is bounded for large enough $\sigma$ and not $o(1)$ there which the $L$ functions satisfy – Conrad Jan 16 '25 at 23:14

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Yes, these bounds both hold, and in fact stronger bounds hold in the former case.

For the former, one uses the trivial bound $L(1 + \varepsilon,\chi) \ll_{\varepsilon} 1$ and the Petrow–Young bound $L(1/2 + it,\chi) \ll_{\varepsilon} (q(1 + |t|))^{1/6 + \varepsilon}$ together with the Phragmén–Lindelöf convexity principle in order to deduce that $L(\sigma + it,\chi) \ll_{\varepsilon} (q(1 + |t|))^{(1 - \sigma)/3 + \varepsilon}$ for $\sigma \in [1/2,1]$.

The latter bound is trivial for $\sigma > 1$. For $\sigma \in [1/2,1]$, it follows by the same method of proof as the case $\sigma = 1/2$, namely the approximate functional equation together with Gallagher's hybrid large sieve.