On a Clarkson-like inequality in $L^p$ I'm currently reading the book "Topics in Almost Everywhere Convergence" by A. Garsia. At a certain point, he claims the validity of the following inequality, for $f, g \in L^p(X, \mathcal{A}, \mu)$, $\mu$ being a probability measure:
\begin{equation*}
\|  f - g \|^{p}_{p} \leq C_{p} \left[ \| f \|_{p}^{p} + \| g \|_{p}^{p} -2 \left\| \frac{f + g}{2} \right\|_{p}^{p} \right]^{\min \{1, p/2\}}
\end{equation*}
for any $p>1$, where $C_p$ is a constant and $f, g$ are on the unit ball. 
If $p\geq 2$, I see that this is Clarkson's inequality. However, for $1 < p <2$, it doesn't seem to be that simple. What bugs me the most is that this result is accompanied by the following footnote:

This can be established by expressing $$|f|^p + |g|^p - 2 \left| \frac{f + g}{2} \right|^p$$ as an integral involving the second derivative of $|x|^p$. The constant $C_p$ tends to infinity as $p \to 1$.

The above suggests that there is some unified approach for all values of $p$ and possibly elementary. Any hint on that direction would be appreciated.
 A: First prove the inequality for positive $f$ and $g$. Begin by obtaining a pointwise inequality, so suppose $f$, $g$ are positive real numbers and that $|f|\le |g|$ and let $\phi(t)=|t|^p$. Then write
\begin{align}
|f|^p+|g|^p-2\left|\frac{f+g}{2}\right|^p& =\left(|g|^p-\left|\frac{f+g}{2}\right|^p\right)-\left(\left|\frac{f+g}{2}\right|^p-|f|^p\right)\\
&= \int\limits_{\frac{|f+g|}{2}}^{|g|} \phi'(t)dt-\int\limits_{|f|}^{\frac{|f+g|}{2}}\phi'(t)dt=\int\limits_{|f|}^{\frac{|f+g|}{2}} (\phi'(t+h)-\phi'(t))dt=\int\limits_{|f|}^{\frac{|f+g|}{2}}\phi''(\xi_{t,h})h \,dt
\end{align}
where $h=\frac{|g-f|}{2}$ and $|f|<\xi_{t,h}<\frac{|f+g|}{2}$ is given by Lagrange.
Since $\phi''(t)=p(p-1)t^{p-2}$, if $1<p\leq 2$ then $\phi''$ is decreasing and therefore $\phi''(\xi_{t,h})\geq \phi''\left(\frac{|f+g|}{2}\right)$; if instead $p\ge 2$ then $\phi''$ is increasing and therefore $\phi''(\xi_{t,h})\geq \phi''(|f|)$. Supposing $1<p<2$ we obtain
$$|f|^p+|g|^p-2\left|\frac{f+g}{2}\right|^p\geq \left(\frac{|f-g|}{2}\right)^2p(p-1)\left(\frac{|f+g|}{2}\right)^{p-2}.$$
Now isolate $(|f-g|/2)^2$, take the power $p/2$ and integrate in $d\mu$ and then use Holder with exponents $\frac2p$ and $\frac{2}{2-p}$ to obtain
$$\left\|\frac{f-g}{2}\right\|_p^p\leq \frac{1}{(p(p-1))^{p/2}}\left(\int |f|^p+|g|^p-2\left|\frac{f+g}{2}\right|^p\right)^{\frac{p}{2}}\left(\int\left|\frac{f+g}{2}\right|^p\right)^{\frac{2-p}{2}}$$
and from
$$\int\left|\frac{f+g}{2}\right|^p\leq \frac12(\|f\|^p+\|g\|^p)\leq 1$$
we conclude finding $C_p=\frac{2^p}{(p(p-1))^{p/2}}$.
If instead $p\geq 2$, use $\phi''(|f|\wedge |g|)$ in the estimate to obtain $C_p=\frac{2^{p+\frac{2-p}{2p}}}{(p(p-1))^{p/2}}$. (EDIT: this is not the right value for $p>2$; in that case we need to get rid of the power $p/2$, raising the RHS to the power $2/p$. This should not be a problem since $f$ and $g$ are in the unit ball, and on bounded sets $|x|\lesssim |x|^\gamma$ if $\gamma<1$).
To prove the inequality for general $f$ and $g$, split them in positive and negative part and apply the triangle inequality together with the inequality just obtained for positive functions, possibly obtaining a worse constant. Maybe there's a way to prove it directly for general $f$ and $g$ but I get a problem because $|f+g|/2$ is not necessarily between $|f|$ and $|g|$.
