Uniform convexity of equivalent intersection norm I have two uniformly convex Banach spaces $E$ and $F$ (which are continuously embedded into a topological vector space $X$) whose intersection $E \cap F$ is non-trivial and is equipped with the norm $\|x\|_{E\cap F} = \sqrt{\|x\|_E^2 + \|x\|_F^2}$. I am having a lot of trouble showing that this norm actually is uniformly convex.
If one assumes that $\|x\|_{E\cap F}, \|y\|_{E\cap F} \leq 1$, we also have $\|x\|, \|y\| \leq 1$ in both the $E$ and $F$ norms. Then if $\varepsilon > 0$ and we assume
$$
\|x-y\|_{E\cap F}^2 = \|x-y\|_E^2+\|x-y\|_F^2 \geq \varepsilon^2\, ,
$$
then we can wlog say that $\|x-y\|_E^2 \geq \frac{\varepsilon^2}{2}$. So by the uniform convexity of $E$ we can write
$$
\left\|\frac{x+y}2\right\|_{E\cap F}^2 = \left\|\frac{x+y}2\right\|_E^2+\left\|\frac{x+y}2\right\|_F^2 \leq (1-\delta)^2 + \left\|\frac{x+y}2\right\|_F^2 \, ,
$$
but then I am more or less stuck. I have tried using the Clarkson inequality
$$
\left | \frac{a+b}2\right |^p + \left | \frac{a-b}2\right |^p\leq \frac 12\left(|a|^p + |b|^p\right) \, ,
$$
for $a,b\in \mathbb{R}$ and $p \in [2,\infty)$ to write
$$
\left\|\frac{x+y}2\right\|_F^2 \leq \left(\frac{\|x\|_F + \|y\|_F}2\right)^2 \leq \frac 12 \left(\|x\|_F^2 + \|y\|_F^2\right)-\left(\frac{\|x\|_F - \|y\|_F}2\right)^2 \leq 1 -\delta' \, ,
$$
but then I get something like
$$
\left\|\frac{x+y}2\right\|_{E\cap F}^2 \leq (1-\delta)^2 + 1-\delta' \, .
$$
By the way, why can't one write
\begin{align*}\left\|\frac{x+y}2\right\|_{E\cap F}^2 &\leq \left(\frac{\|x\|_{E\cap F} + \|y\|_F}2\right)^2 \leq \frac 12 \left(\|x\|_{E\cap F}^2 + \|y\|_{E\cap F}^2\right)-\left(\frac{\|x\|_{E\cap F} - \|y\|_{E\cap F}}2\right)^2 \\
&\leq 1 -\delta\, ,\end{align*}
directly? Obviously, this can't be used to show uniform convexity since we would not be using the uniform convexity of neither $E$ nor $F$, but I don't see the fault in this argument.
Can anyone point me in the right direction regarding this problem? Thanks.
 A: The result from problem 3.29 from Brezis' Functional Analysis, says the following.

If $E$ is a uniformly convex Banach space. Then $\forall \varepsilon > 0, M > 0 \exists \delta > 0$ we have
  $$\|x-y\| > \varepsilon \Longrightarrow \left\|\frac{x+y}{2}\right\| \leq \frac 12\left(\|x\|^2 + \|y\|^2\right)-\delta \, ,$$
  for every $x,y\in E$ with $\|x\|, \|y\| \leq M$.

If we assume that $x,y\in E \cap F$ with $\|x\|_{E\cap F}, \|y\|_{E\cap F} \leq 1$. This implies $\|x\|_E^2+\|x\|_F^2 \leq 1$ and $\|y\|^2_E+\|y\|_F^2 \leq 1$. Assuming that $\|x-y\|_{E\cap F}^2 = \|x-y\|_E^2 + \|x-y\|_F^2 > \varepsilon$ we have $x\neq y$ in $E\cap F$ and hence $x\neq y$ in both $E$ and $F$. I then believe we can draw the conclusion that we without loss of generality can assume that $\|x-y\|_E^2 > \varepsilon'$ and $\|x-y\|_F^2 > \varepsilon''$ for any $\varepsilon', \varepsilon'' > 0$. In this case we can use the above result to get the following.
\begin{align*}
\left\|\frac{x+y}2\right\|_{E\cap F}^2 &= \left\|\frac{x+y}2\right\|_E^2+\left\|\frac{x+y}2\right\|_F^2 \leq \frac 12 \left(\|x\|_E^2 + \|x\|_F^2 + \|y\|_E^2 + \|y\|_F^2\right) - \delta'-\delta'' \\
&\leq 1-\delta'-\delta''\, ,
\end{align*}
for $\delta', \delta'' > 0$, and hence $E\cap F$ is also uniformly convex.
