# 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

• This is not Lemma 3.6 in that paper, but Lemma 5.4 not in that paper since it stops at section 4. The proof of Lemma 3.6 is different. Commented Oct 13, 2021 at 5:07
• @username I have corrected the wrong description , and these two papers are the same author. Commented Oct 13, 2021 at 6:35

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 $$\|v^-\|_2 \leq C\epsilon^{7/4}$$.

• This is the right answer. But it's worth noting that all counterexamples are "like" this one, in the sense that they have $(u, \varphi_1) \geq c \|u\|_{L^2}$. Commented Oct 12, 2021 at 21:36
• @user378654 Could you explain with some more detail what you mean? Many thanks Commented Oct 13, 2021 at 6:29

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 $$\varphi_j$$.

• Many thanks for the explanation. How do you see that the right minimizer is what you indicated? Commented Oct 13, 2021 at 13:15
• Well, to minimize the integral you want to put all the mass where $\varphi_1$ is smallest. You can justify formally using the layercake integration formula $\int \varphi_1 |u| = \int_0^\infty t\int_{\{ \varphi_1 = t\}} |u(x)| d\mathcal{H}^{n-1}(x) dt$. Commented Oct 13, 2021 at 17:05