A property of eigenfunctions of p-Laplacian in $\mathbb{R}^d$ when $p=d+1$ Consider the $(d+1)$-Laplacian denoted as $\Delta_{d+1}$, ($p$-laplacian with $p=d+1$) and its eigenfunctions in $\mathbb{R}^d$. 
I'd like to know whether we can say that any linear combination of eigenfunctions belonging to the same eigenvalue is also an eigenfunction with the same eigenvalue? ( despite the fact that not all eigenfunctions of same eigenvalue are necessarily scaled versions of one another). It is easily verified for the case $d=1$.
PS : 
A function $u$ is eigen function corresponding to eigenvalue $\lambda$, if it satisfies the following equation
$$ 0= \lambda   u |u|^{d-1} +\nabla \cdot( \nabla u |\nabla u|^{d-1}) $$
 A: The desired properly fails when $n\geq2$ and $\lambda=0$.
Probably it also fails for $\lambda\neq0$, but the special case is enough to show that the claim is false as stated.
If the eigenvalue is zero, the eigenfunctions are simply the $p$-harmonic functions.
It is not important that $p=d+1$; the only thing that matters is $p\neq2$.
Then the question is whether the sum of two $p$-harmonic functions is always $p$-harmonic.
The answer is no, and there are probably several ways to see it.
Here is one method chosen at random.
Perhaps the simplest way would be to choose two explicit $p$-harmonic functions and show that the sum is not $p$-harmonic.
(One could argue that since the equation is not linear, the space of solutions shouldn't be linear, either. But that's not a sound argument as such.)
Suppose $u$ and $v$ are $p$-harmonic (and in $W^{1,p}_{\text{loc}}(\mathbb R^d)$) and the sum of any two such functions is also $p$-harmonic.
Let $f_h=u+hv$ for any $h\in\mathbb R$.
Assume that $u$ is not constant.
For convenience, consider the equation in a small open set $\Omega$ where $|\nabla u|$ is bounded away from zero.
Such a set exists due to local regularity results unless $u$ is constant.
We have that
$$
\begin{split}
0
&=
\Delta_p f_h
\\&=
\nabla\cdot(|\nabla f_h|^{p-2}\nabla f_h)
\\&=
\nabla\cdot[|\nabla u+h\nabla v|^{p-2}(\nabla u+h\nabla v)]
\\&=
\nabla\cdot[|\nabla u|^{p-2}(1+(p-2)h|\nabla u|^{-2}\nabla u\cdot\nabla v+O(h^2))(\nabla u+h\nabla v)]
\\&=
\Delta_pu
+
h
\nabla\cdot[
|\nabla u|^{p-2}
((p-2)|\nabla u|^{-2}\nabla u\cdot\nabla v)\nabla u
+
\nabla v)
]
+
O(h^2).
\end{split}
$$
This vanishes for all $h$, so
$$
\nabla\cdot[
|\nabla u|^{p-2}
((p-2)|\nabla u|^{-2}\nabla u\cdot\nabla v)\nabla u
+
\nabla v)
]
=
0.
$$
If we now choose $u(x)=x_i$, we get
$$
(p-2)\partial_i^2 v
+
\Delta v
=
0.
$$
Summing this over all $i$ gives $(p+d-2)\Delta v=0$, implying that $\Delta v=0$.
That is, we have concluded that any $p$-harmonic function $v$ is also harmonic.
But this is not true.
To show this, simply examine your favorite example of a non-affine $p$-harmonic function.
A: To get started, and to throw more light on what I am seeking, I'd like to verify for the case $d=1$, that is $\mathbb{R}$.
One can verify the equation in $\mathbb{R}$ is $$u''+\lambda u = 0$$ and any function of the form $$u(x) = A\sin(\sqrt{\lambda} x + \theta), \forall A,\theta \in \mathbb{R}$$ is an eigenfunction (Please note that there are no boundary or any boundary restrictions in the problem considered in this Q&A). Note that not all these eigenfunctions are scaled versions of one another.
Proof basis: $$A_1\sin(\sqrt{\lambda} x + \theta_1) + A_2\sin(\sqrt{\lambda} x + \theta_2) = A_3\sin(\sqrt{\lambda} x + \theta_3)$$
