A little problem in PDE or function analysis Let

*

*$E$ be the usual sobolev space $H^{1}_{0}(\Omega)$ on a smoothly bounded domain $\Omega$,

*$E_{k}$ be its subspace spanned by the first $k$ eigenfunctions of the Laplace operator, i.e.
$$E_{k}:=\text{span}\{\varphi_{j}\in E: -\Delta\varphi_{j}=\lambda_{j}\varphi_{j},~j=1,2\dots,k \},$$

*$P$ be the positive cone in $E$, i.e,
$$P:=\{u \in E~; u \geq 0 ~a.e.\},$$
Now set $P_{k}=P\cap E_{k}~~u^{-}=min\{u,0\}$.
My question: does it exist $C_{k}>0$, such that
$$\text{dist}_{L^{2}}(u,P_{k}) \leq  C_{k} \text{dist}_{L^{2}}(u,P)~\text{ holds }~\forall u\in E_{k}\; ?$$
or
$$\text{dist}_{E}(u,P_{k}) \leq  C_{k} \text{dist}_{E}(u,P)~\text{ holds }~\forall u\in E_{k}\; ?$$
or $$\text{dist}_{E}(u,P_{k}) \leq  C_{k} \|u^{-}\|_{E}~\text{ holds }~\forall u\in E_{k}\; ?$$
If not, could you please show me a counterexample? Thanks.
Moreover, this problem is raised from the proof of lemma 3.6 in "Infinitely many solutions to perturbed elliptic equations",doi:10.1016/j.jfa.2005.06.014
enter image description here
also see lemma 5.4 in "On finding sign-changing solutions",doi:10.1016/j.jfa.2005.09.004

 A: The first is not true, and probably also the others.
Take $L^2(0, \pi)$ and $u_1=\sin x$, $u_2=\sin (2x)$, so that $E_2=\{u=a\sin x+b \sin (2x)\}$ and $u \geq 0$ iff $a \geq 0$ and $2|b| \leq a$. If $v=\alpha sin x+\beta \sin (2x)$, then $\|u-v\|_2^2=\frac{\pi}{2} \left((a-\alpha)^2+(b-\beta)^2\right)$ and, if $v_\epsilon=\sin x-\frac{1+\epsilon}{2}\sin (2x)$, the closest positive $u \in E_2$ is $u=\sin x-\frac{1}{2}\sin (2x)$ and $\|v_\epsilon -u\|_2 \approx \epsilon$. On the other hand, $v_\epsilon$ is negative in an interval starting from 0 of length $\approx \sqrt \epsilon$ where the function is of order $\epsilon^{3/2}$ and $\|v^-\|_2 \leq C\epsilon^{7/4}$.
A: This does not answer the question asked (see the other answer for a good counterexample) and I don't know if it's relevant to the paper. However, if $u$ is sufficiently far from $\varphi_1$, the first eigenfucntion, you do get a positive answer. Below I assume $\Omega$ is connected (you'd have to work out what happens more carefully if it's not).
First, we have that
$$
\|u\|_{L^\infty} \leq C(k) \|u\|_{L^2}
$$
for any $u \in E_k$; this can be checked for each $\phi_j$ and is standard (there is a clean argument with the heat kernel, or you can just apply the local maximum principle repeatedly on balls). Also, $\varphi_1 > 0$ on $\Omega$.
Now consider $u \in E_k$ with $\int u^2 = 1$ and
$$
\int u \varphi_1 = a.
$$
We have that
$$
\int |u|\varphi_1 \geq c(k)
$$
using only that $|u|\leq C(k)$ and $\int u^2 = 1$. Indeed, this integral is minimized by a piecewise constant function $u$ which is $C(k)$ on $F = \{\varphi_1 < t\}$ and $0$ outside of $F$, where $t$ is chosen so that $|F| = 1/C(k)^2$. Setting $c(k) = C(k) \int_F \varphi_1 > 0$ gives the inequality.
Combining, we have that
$$
\|u_-\|_{L^2} \geq \int u_- \varphi_1 \geq \frac{c(k) - a}{2}.
$$
Undoing the normalization, we have shown that for $u \in E_k$, if $\int u \varphi_1 \leq \frac{c(k)}{2} \|u\|_{L^2}$, then
$$
\|u\|_{L^2} \leq \frac{4}{c(k)} \|u_-\|_{L^2}.
$$
This implies both inequalities in the question in this case (the $L^2$ and $H^1_0$ norms of each side are comparable). In particular, it applies to any $u \in E_k$ orthogonal to $\varphi_1$, so to any eigenfunction $\varphi_j$.
