Let $K,L$ be closed subspaces of a Hilbert space $H$, Prove that $\|P_K+P_L\|=1$ iff $K$ and $L$ are orthogonal
My first observation is that for forwards direction $K\cap L=\{0\}$ otherwise the norm would be at least $2$.
If we assume towards contradiciton they are not orthogonal complements, we end up with an elements $x$ s.t $x\in K$ and $x\not \in L^{\perp}$. Thus we get $\|P_K(x)+P_L(x)\|=\|x+P_L(x)\|$ i am not sure how to show this can have norm greater than $1$.
Best Answer
Assume that $K$ and $L$ are orthogonal. Let $\mathcal{K}, \mathcal{L}$ be orthonormal Bases of $K$ and $L$. We can extend them to an orthonormal Basis $\mathcal{H}$ of $H$. (Note that $\mathcal{H}$ is countable iff $H$ is separable.) Thus \begin{align} \|P_K + P_L\|^2 = \sup_{h\in H, \|h\|=1} \|P_K(h) + P_L(h)\|^2 = \sup_{\|h\|=1} \|P_K(h)\|^2 + \|P_L(h)\|^2\\ = \sup_{\|h\|=1} \sum_{k \in \mathcal{K}} |\langle h , k\rangle|^2 + \sum_{l \in \mathcal{L}} |\langle h , l\rangle|^2 \leq \sup_{\|h\|=1} \|h\|^2 = 1\,. \end{align} Furthermore for $k \in K$ st. $\|k\|=1$ we have $$\|P_K+P_L\| \geq \|P_K(k)+P_L(k)\| = \|k\|= 1\,.$$
On the contrary assume that there is an $x∈K$ st. $\|x\|=1$ and $x∉L^⊥$. Note that $H=L\oplus L^⊥$ and therefore $x = P_L(x) + P_{L^\perp}(x)$. It follows that \begin{align} \|P_K(x)+P_L(x)\|^2 = \|x+P_L(x)\|^2 &= \langle x , x \rangle + 2 Re(\langle x , P_L(x) \rangle) +\langle P_L(x) , P_L(x) \rangle\\ &=\|x\|^2+3\|P_L(x)\|^2 >1 \end{align}