TristanvanLeeuwen · SCdummer · Feb 23, 2024
diff --git a/ip_function_spaces.md b/ip_function_spaces.md
@@ -114,7 +114,7 @@ Next, we show that the normal equations only allow a solution when $f \in \mathc
 
 * Given a solution $\widetilde{u}$, we find that $f = K\widetilde{u} + g$ with $g\in\mathcal{R}(K)^\perp$. Hence, $f \in \mathcal{R}(K) \oplus \mathcal{R}(K)^\perp$. Conversely, given $f \in \mathcal{R}(K) \oplus \mathcal{R}(K)^\perp$ , there exist $u \in \mathcal{U}$ and $g \in \mathcal{R}(K)^\perp = \left(\overline{\mathcal{R}(K)}\right)^\perp$ such that $f = Ku + g$. Thus, $P_{\overline{\mathcal{R}(K)}}f = Ku$. Such a $u \in \mathcal{U}$ must necessarily be a solution to {eq}`minres` because we have $\|Ku - f\|_{\mathcal{F}} = \|P_{\overline{\mathcal{R}(K)}}f - f\|_{\mathcal{F}} \leq \inf_{g\in \overline{\mathcal{R}(K)}}\|g  - f\|_{\mathcal{F}} \leq \inf_{v\in\mathcal{U}}\|Kv - f\|_{\mathcal{F}}.$ Here, we used the fact that the orthogonal projection on a subspace gives the element closest to the projected element and $\mathcal{R}(K) \subseteq \overline{\mathcal{R}(K)}$ allows to conclude the last inequality.
 
-Finally, we show that the minimum-norm solution is unique. Denote the minimun-norm solution by $\widetilde{u}$. Now suppose we have another solution, $\widetilde{v}$, to {eq}`minres`. Since they both solve the normal equations we have $\widetilde{v} = \widetilde{u} + z$ with $z \in \mathcal{N}(K)$. It follows that $\|\widetilde{v}\|_{\mathcal{U}}^2 = \|\widetilde{u}\|_{\mathcal{U}}^2 + 2\langle \widetilde{u}, z \rangle_{\mathcal{U}} + \|z\|_{\mathcal{U}}^2$. Since $\widetilde{u} \in \mathcal{N}(K)^\perp$ we have  $\langle \widetilde{u}, z \rangle_{\mathcal{U}} = 0$ and hence $\|\widetilde{v}\|_{\mathcal{U}} \geq \|\widetilde{u}\|_{\mathcal{U}}$ with equality only obtained when $z = 0$.
+Finally, we show that the minimum-norm solution is unique. Denote the minimun-norm solution by $\widetilde{u}$. As $\mathcal{U} = \mathcal{N}(K)^\perp \oplus \mathcal{N}(K)},$, we can write $\widetilde{u} = v + z$ with $v \in \mathcal{N}(K)^\perp$ and $z \in \mathcal{N}(K)$. It follows that $\|\widetilde{u}\|_{\mathcal{U}}^2 = \|v\|_{\mathcal{U}}^2 + 2\langle v, z \rangle_{\mathcal{U}} + \|z\|_{\mathcal{U}}^2$. Since $v \in \mathcal{N}(K)^\perp$ and $z \in \mathcal{N}(K)$ we have  $\langle v, z \rangle_{\mathcal{U}} = 0$ and hence $\|\widetilde{u}\|_{\mathcal{U}} \geq \|v\|_{\mathcal{U}}$ with equality only obtained when $z = 0$. So the unique minimum norm solution is $\widetilde{u} = v \in \mathcal{N}(K)^\perp$.
 
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     * Given a solution $\widetilde{u}$, we find that $f = K\widetilde{u} + g$ with $g\in\mathcal{R}(K)^\perp$. Hence, $f \in \mathcal{R}(K) \oplus \mathcal{R}(K)^\perp$. Conversely, given $f \in \mathcal{R}(K) \oplus \mathcal{R}(K)^\perp$ , there exist $u \in \mathcal{U}$ and $g \in \mathcal{R}(K)^\perp = \left(\overline{\mathcal{R}(K)}\right)^\perp$ such that $f = Ku + g$. Thus, $P_{\overline{\mathcal{R}(K)}}f = Ku$. Such a $u \in \mathcal{U}$ must necessarily be a solution to {eq}`minres` because we have $\|Ku - f\|_{\mathcal{F}} = \|P_{\overline{\mathcal{R}(K)}}f - f\|_{\mathcal{F}} \leq \inf_{g\in \overline{\mathcal{R}(K)}}\|g  - f\|_{\mathcal{F}} \leq \inf_{v\in\mathcal{U}}\|Kv - f\|_{\mathcal{F}}.$ Here, we used the fact that the orthogonal projection on a subspace gives the element closest to the projected element and $\mathcal{R}(K) \subseteq \overline{\mathcal{R}(K)}$ allows to conclude the last inequality.
-    Finally, we show that the minimum-norm solution is unique. Denote the minimun-norm solution by $\widetilde{u}$. Now suppose we have another solution, $\widetilde{v}$, to {eq}`minres`. Since they both solve the normal equations we have $\widetilde{v} = \widetilde{u} + z$ with $z \in \mathcal{N}(K)$. It follows that $\|\widetilde{v}\|_{\mathcal{U}}^2 = \|\widetilde{u}\|_{\mathcal{U}}^2 + 2\langle \widetilde{u}, z \rangle_{\mathcal{U}} + \|z\|_{\mathcal{U}}^2$. Since $\widetilde{u} \in \mathcal{N}(K)^\perp$ we have  $\langle \widetilde{u}, z \rangle_{\mathcal{U}} = 0$ and hence $\|\widetilde{v}\|_{\mathcal{U}} \geq \|\widetilde{u}\|_{\mathcal{U}}$ with equality only obtained when $z = 0$.
+    Finally, we show that the minimum-norm solution is unique. Denote the minimun-norm solution by $\widetilde{u}$. As $\mathcal{U} = \mathcal{N}(K)^\perp \oplus \mathcal{N}(K)},$, we can write $\widetilde{u} = v + z$ with $v \in \mathcal{N}(K)^\perp$ and $z \in \mathcal{N}(K)$. It follows that $\|\widetilde{u}\|_{\mathcal{U}}^2 = \|v\|_{\mathcal{U}}^2 + 2\langle v, z \rangle_{\mathcal{U}} + \|z\|_{\mathcal{U}}^2$. Since $v \in \mathcal{N}(K)^\perp$ and $z \in \mathcal{N}(K)$ we have  $\langle v, z \rangle_{\mathcal{U}} = 0$ and hence $\|\widetilde{u}\|_{\mathcal{U}} \geq \|v\|_{\mathcal{U}}$ with equality only obtained when $z = 0$. So the unique minimum norm solution is $\widetilde{u} = v \in \mathcal{N}(K)^\perp$.
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