Is a bounded linear operator $L$ from a Hilbert space $U$ to a Hilbert space $H$ continuous from $U$ to $LU$? Let


*

*$U,H$ be $\mathbb R$-Hilbert spaces

*$L$ be a bounded linear operator from $U$ to $H$

*$LU$ be equipped with $$\langle u,v\rangle_{LU}:=\langle L^{-1}u,L^{-1}v\rangle_U\;\;\;\text{for }u,v\in LU\;,$$ where $L^{-1}:=\left(\left.L\right|_{\left(\ker Q\right)^\perp}\right)^{-1}$ denotes the pseudo inverse of $L$ (Note that $LU$ is a Hilbert space and $L^{-1}$ is an isometry from $LU$ to $U$)



How can we show that $L$ is continuous from $U$ to $LU$?

Let $u\in U$. Since $U=\ker L\oplus(\ker L)^\perp$, there are $u_0\in\ker L$ and $v\in(\ker L)^\perp$ with $$u=u_0+v$$ and hence $$\left\|Lu\right\|_{LU}=\left\|L^{-1}(Lu)\right\|_U=\left\|L^{-1}(Lv)\right\|_U=\left\|v\right\|_U\tag1\;.$$ If $L$ would be continuous, we would have $\left\|Lu\right\|_{LU}\le\left\|u\right\|_U$ and hence $$\left\|v\right\|_U\le\left\|u\right\|_U\tag2$$ by $(1)$.

$(2)$ seems to be wrong. I've found the claim in a lecture note. Are we able to  prove it or is the claim wrong?


The statement $(2)$ is correct. Since $v \perp u_0$, the Pythagorean identity gives:
$$\|v\|_U^2 \le \|v\|_U^2 + \|u_0\|_U^2 =  \|v+u_0\|_U^2 = \|u\|_U^2$$
So $\|v\|_U \le \|u\|_U$.
Therefore, $$\|Lu\|_{LU} = \|v\|_U \le \|u\|_U$$
so $L$ is bounded and $\|L\| \le 1$. We conclude that $L$ is continuous.
