Is every orthogonal projection continuous?

Question

Is there an example of a non-continuous linear operator $\pi$ on a $\mathbb R$-Hilbert space $H$ with $\pi^2=\pi$ and orthogonal null space and range?

Clearly, if the range is closed, then $\pi$ is continuous.

Answer 1

Every orthogonal projection $\pi$ satisfies $\|\pi(x)\|\le \|x\|$ for every $x\in H$. We can observe that for every $x\in H$, $$ x =\pi(x) + \left(x-\pi(x)\right). $$ We find that $\pi(x) \in \text{im}\ \pi$ and $x-\pi(x)\in \ker \pi$ since $\pi$ is an idempotent. Thus they are orthogonal to each other, and from this, we know that $$ \|x\|^2 = \|\pi(x)\|^2 +\|x-\pi(x)\|^2 \ge \|\pi(x)\|^2, $$ i.e. $\|\pi(x)\|\le \|x\|$. This gives $\|\pi\|\le 1$, hence every orthogonal projection has a norm at most $1$. (If it has non-trivial range, we can show $\|\pi\|=1$ holds actually.)