Continuity of $L_p$ norm in $p$ with $\varepsilon$-$\delta$ definition Assume that $\|f\|_p< \infty$ for $1\le p<\infty$.
In this question we showed that
$$
g(p)=\|f\|_p
$$
is continuous in $p \ge 1$. The technique was to use Dominant Convergence theorem.
Using $\varepsilon$-$\delta$ language, what this means is that for any $\varepsilon>0$ there is a $\delta>0$ such that for all $|q-p| < \delta(\varepsilon)$ implies that
$$
\left | \|f\|_p-\|f\|_q \right| \le \varepsilon
$$
My question the following. Can we characterize $\delta(\varepsilon)$ more explicitly in term of $\varepsilon$ and have an expression for $\delta$?
Observer, that $\delta$ should probably be a function of $p$ as well, otherwise I don't think it is possible.
 A: Here's a super-soft answer. Fix a measurable function $f$ such that $f\in L^p$ for all $p\in (p_-, p_+)$ ($p_+$ possibly being $\infty$). Let 
$$\Phi\left(\frac 1 p\right)=\left[ \int \lvert f\rvert^p\right]^\frac{1}{p}.$$
This function $\Phi$ is log-convex on the interval $\left(\frac1{p_+}, \frac1{p_-}\right)$, meaning that it satisfies the following inequality:
$$
\Phi\left( (1-\alpha)\frac1{p_1} + \alpha \frac1{p_2}\right)\le \Phi\left(\frac1{p_1}\right)^{1-\alpha}\Phi\left(\frac{1}{p_2}\right)^{\alpha}$$
where $p_1, p_2\in (p_-, p_+)$ and $\alpha\in [0, 1]$. (This inequality is a consequence of Hölder's inequality and it gives an alternative proof of the continuity of $\|f\|_p$ with respect to $p$). 
Now any log-convex function is convex and any convex function is Lipschitz on compact subintervals of its interval of definition (one says that it is locally Lipschitz). So $\Phi$ is locally Lipschitz on $\left(\frac1{p_+}, \frac1{p_-}\right)$, which means that
$$
\left\lvert \|f\|_{L^{p_1}} - \|f\|_{L^{p_2}}\right\rvert \le C_{f, I}\left\lvert \frac1{p_1} -\frac1{p_2}\right\rvert,\qquad \forall p_1, p_2\in I$$
where $I\subset (p_-, p_+)$ is a compact interval. 
