# Characterization on smallest element in affine Sobolev subspace

Suppose we are given a sequence $$\phi_k$$ of traces (i.e. functions defined on boundary $$\partial B_1$$) such that $$\phi_k \rightarrow 0 \;\mbox{in L^{\infty}(\partial B_1)}$$ (one can consider $$C^{\alpha}$$ convergence if required).

Now we consider affine subspaces Sobolev space $$W^{1,p}(B_1)$$ $$W^{1,p}_{\phi_k}(B_1):= \Big \{ v\in W^{1,p}(B_1): \text{Trace}(v) =\phi_k \Big \}$$ Let $$v_k\in W^{1,p}_{\phi_k}(B_1)$$ be such that $$v_k$$ has least $$W^{1,p}(B_1)$$ norm in the space $$W^{1,p}_{\phi_k}(B_1)$$. That is $$v_k$$ is the minimizer of the following convex functional $$J_k(v):= \int_{B_1}(|\nabla v|^p +|v|^p)\,dx,\;\;v\in W^{1,p}_{\phi_k}(B_1).$$ We can see that $$v_k$$ satisfies the following Euler-Lagrange equation of $$J_k$$ : $$\begin{cases} -\Delta_p v_k = |v_k|^{p-2}v_k\;\;\mbox{in B_1}\\ v_k =\phi_k\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mbox{ on \partial B_1}. \end{cases}$$

Question is since the traces $$\phi_k$$ are tending to Zero in $$L^{\infty}(\partial B_1)$$, is there any result which shows that the smallest elements in $$W^{1,p}_{\phi_k}(B_1)$$ also tend to zero in strong $$W^{1,p}(B_1)$$ topology? We have to show that $$v_k \rightarrow 0 \;\;\mbox{in W^{1,p}(B_1)}.$$

Isn't it the case that the smallest possible Sobolev norm for an element in $$W^{1,p}_{\phi_k}(B_1)$$ has to be controlled by some norm of $$\phi_k$$.

The above question can also be seen in terms of $$\Gamma$$ convergence where we try to show that $$J_k \xrightarrow{\Gamma} 0$$ in weak $$W^{1,p}(B_1)$$ topology.

Moreover, we can also see it as eigenvalue problem for $$p$$-Laplacian.

• i think you have a sign wrong in the equation Jul 3, 2021 at 18:00

It seems to depend on the strength assumed of the convergence of the $$\phi_k$$. This is a partial answer, where in some parts it is assumed that $$p > n - 1$$. Under this hypothesis, the convergence does not hold if only $$\lvert \phi_k \rvert_{L^\infty(\partial B_1)} \to 0$$. However if $$\lvert \phi_k \rvert_{C^{0,\beta}(\partial B_1)} \to 0$$ for some $$\beta \in (1 - \frac{1}{p},1)$$ then also $$\lvert v_k \rvert_{W^{1,p}(B_1)} \to 0$$, regardless of the relative size of $$p$$ and $$n$$.
The trace operator is instrumental in seeing this: recall that there is a constant $$C > 0$$ so that $$\lvert \mathrm{tr} \, u \rvert_{W^{1-1/p,p}(\partial B_1)} \leq C \lvert u \rvert_{W^{1,p}(B_1)}$$ for all $$u \in W^{1,p}(B_1)$$. Here and throughout $$W^{1-1/p,p}(\partial B_1)$$ is a fractional Sobolev-Slobodeckij space. Recall also that there is an extension operator $$W^{1,1-1/p}(\partial B_1) \to W^{1,p}(B_1)$$ that is a right inverse to the trace.
In particular, if there were the strong convergence $$\lvert v_k \rvert_{W^{1,p}(B_1)} \to 0$$ then also $$\lvert \phi_k \rvert_{W^{1-p,p}(\partial B_1)} \to 0.$$ This is strictly stronger than $$L^\infty$$-convergence because $$W^{1-1/p,p}(\partial B_1)$$ embeds into the Holder space $$C^{0,\alpha}(\partial B_1)$$, where $$\alpha \in (0,1 - \frac{1}{p} -\frac{n-1}{p}] = (0,1-\frac{n}{p}]$$. Therefore, if one chose a sequence of traces so that $$\lvert \phi_k \rvert_{C^{0,\alpha}(\partial B_1)} \not\to 0$$ for some exponent $$\alpha$$ in this range then the convergence $$\lvert v_k \rvert_{W^{1,p}(B_1)} \to 0$$ would be impossible. (This is where the hypothesis $$p > n - 1$$ is used: when $$p < n - 1$$ then instead $$W^{1-p,p}(\partial B_1) \subset L^s(\partial B_1)$$ for all $$s \in (0,\frac{(n-1)p}{n-1-p}]$$.)
However, if $$\beta > 1 - \frac{1}{p}$$ and one assumes $$\lvert \phi_k \rvert_{C^{0,\beta}(\partial B_1)} \to 0$$ then the convergence $$\lvert v_k \rvert_{W^{1,p}(B_1)} \to 0$$ is guaranteed. This is because the 'reverse' inclusion $$C^{0,\beta}(\partial B_1) \subset W^{1-1/p,p}(\partial B_1)$$ holds. Therefore $$\phi_k \in W^{1-1/p,p}(\partial B_1)$$; let $$u_k \in W^{1,p}(B_1)$$ be its image under the extension operator. By the above there is $$C > 0$$ so that $$\lvert u_k \rvert_{W^{1,p}(B_1)} \leq C \lvert \phi_k \rvert_{C^{0,\beta}(\partial B_1)} \quad \text{for all k}.$$ By minimality, the same holds with $$v_k$$ replacing $$u_k$$, and therefore $$\lvert v_k \rvert_{W^{1,p}(B_1)} \to 0$$ also. (Note that the assumption that $$p > n-1$$ is not needed here.)