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.
Thank  you in advance.
 A: 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.)
