Let $\Omega\subset\mathbb{R}^n$ be a bounded domain and $p\in (1,\infty)$, $p\neq 2$. Consider the usual Sobolev space $W_0^{1,p}(\Omega)$ and its dual $W^{-1,p'}(\Omega)$, where $1/p+1/p'=1$. Define $\lambda_1>0$ by $$\tag{1}\lambda_1=\inf_{u\in W_0^{1,p}(\Omega)}\frac{\|u\|_{1,p}^p}{\|u\|_p^p}$$
$\lambda_1$ is the first eigenvalue associated with the problem $$-\Delta _p u=\lambda |u|^{p-2}u,\ u\in W_0^{1,p}(\Omega),\tag{2}$$
where the above equation is understood as an equation in $W^{-1,p'}(\Omega)$ and $\Delta_p u=|\nabla u|^{p-2}\nabla u$. As you can see in these notes from Peral, the infimum in $(1)$ is achieved by, let's say, $u_1\in W_0^{1,p}(\Omega)$. Moreover, $u_1$ does not change sign, i.e. we can choose it to be strictly positive in $\Omega$ and if $v$ is another minimum point of $(1)$, we have that $v=\alpha u_1$, $\alpha\in \mathbb{R}$ ($\lambda_1$ is simple).
I would like to note that $(1)$ can also be viewed as a problem of minimization on the Banach manifold $$S_p=\{u\in W_0^{1,p}(\Omega):\ \|u\|_p=1\}$$
i.e. $$\lambda_1=\inf_{u\in W_0^{1,p}(\Omega),\ u\in S_p}\|u\|^p_{1,p}\tag{3}$$
We have from the simplicity of $\lambda_1$ that there are two eigenvectors of $(2)$ in $S_p$. Fix one of them and let's call it $u_1$. The tangent space in the point $u_1$ in $S_p$ can be indetified as $$T=T_{u_1}S_p=\left\{v\in W_0^{1,p}:\ \int_\Omega |u_1|^{p-2}u_1 v=0\right\}$$
Let $T_1=\{v\in T:\ \|v\|_{1,p}=1\}$. Consider the function $F:\mathbb{R}\times T_1\to \mathbb{R}$ defined by $$F(t,v)=\frac{\int_\Omega|\nabla (u_1+tv)|^p}{\int_\Omega|u_1+tv|^p}=\frac{\|u_1+tv\|_{1,p}^p}{\|u_1+tv\|_p^p}=\left\|\frac{u_1+tv}{\|u_1+tv\|_p}\right\|_{1,p}^p$$
I am trying to prove that there exist $\epsilon>0$ such that for all fixed $v\in T_1$ the function $f(t,v)$ is strictly convex for $t\in (-\epsilon,\epsilon)$. Geometrically, I am taking curves in $S_p$ which starts in $u_1$.
If someone has an idea, maybe some trick inequality or even a counter example, I would be grateful.