I'm attempting to read a proof in Taylor's PDE, Vol I. Chapter 5. And I'm stuck on a detail in Proposition 1.7:
Consider the following boundary problem for $u$: $$ \Delta u =0\text{ on }\Omega,\hspace{.5 in}u\big|_{\partial\Omega}=f, $$ where $f\in C^{\infty}(\partial\Omega)$ is given. We denote its solution by $$ \text{PI }f. $$ Proposition 1.7. The map above has a continuous extension $\text{PI }: H^s(\partial\Omega)\to H^{s+1/2}(\Omega)$ for $s\geq 1/2$.
Proof. It suffices to prove this for $s=k+1/2, k=0,1,2,\ldots $ by interpolation. Given $f\in H^{k+1/2}(\partial\Omega)$, there exists $F\in H^{k+1}(\Omega)$ such that $F\big|_{\partial\Omega}=f$, by surjectivity of the trace operator. Then $\text{PI }f=F+v$, where $v$ is given as the solution to $$ \Delta v=-\Delta F\in H^{k-1}(\Omega),\hspace{.5 in} H_0^1(\Omega). $$ by the elliptic theory of the Laplacian, which establishes the result for $s=k+1/2$.
Now, like this question from a few years back (but for different reasons), I don't understand why this operator is continuous. What does make sense to me is that given $f\in H^{k+1/2}(\partial\Omega)$, we have \begin{align*} \|\text{PI }f\|_{H^{k+1}(\Omega)}&\leq \|F\|_{H^{k+1}(\Omega)}+\|v\|_{H^{k+1}(\Omega)}\\ &\lesssim \|F\|_{H^{k+1}(\Omega)}+\|\Delta F\|_{H^{k-1}(\Omega)}\\ &\lesssim \|F\|_{H^{k+1}(\Omega)}. \end{align*} Now from here I'm evidently supposed say that $$ \|F\|_{H^{k+1}(\Omega)}\lesssim\|\text{Trace }(F)\|_{H^{k+1/2}(\partial\Omega)}=\|f\|_{H^{k+1/2}(\partial\Omega)} $$ and be done, but simply don't see why this should be true. I know the trace operator is bounded, but this is asking for the opposite bound. Is there something I'm missing that's so trivial that none of the textbooks I've consulted bother addressing it? I've messed around trying to get this bound via the Fourier transform when $\Omega$ is a half-space, but to no avail. Any help is greatly appreciated. Thanks.
Edit:
This question also seems to share my concern. I've only discovered it after posting.
I only know the case $k=0$, but maybe this will be enough to help for general $k$.
The trace (restriction) operator $R:H^1\to H^{1/2},u\mapsto [u]=\{u+v:v\in H^1_0\}$ has a continuous right inverse $E:H^{1/2}\to H^1$, $R\circ E=1_{H^{1/2}}$. Indeed, since $H^1_0\subset H^1$ is a closed subspace, we have an orthogonal decomposition $H^1\simeq H^1_0\oplus (H^1_0)^\perp$. The complement $(H^1_0)^\perp$ captures the "$H^{1/2}$ part". If we let $P:H^1\to (H^1_0)^\perp\subset H^1$ be the orthogonal projection, then consider the map $E:H^{1/2}\to H^1,[u]\mapsto P(u)$. It is well defined since if $u\in H^1_0$, then $P(u)=0$ and $[u]=[0]$. Since it is a right inverse of $R$, we need only verify its continuity: $$ \|E(u)\|_{H^1}=\|P(u)\|_{H^1}=\|P(u)\|+\inf_{v\in H^1_0}\|v\|_{H^1}= \inf_{v\in H^1_0}\|P(u)+v\|_{H^1}=\|\,[u]\,\|_{H^{1/2}}. $$ Note of course: $P(u)\perp v$, and $P(u)+v=u+(v-\pi_{H^1_0}u)$.