Isometry between $L^{\bot}$ and $H/L$ where $H$ is hilbert and $L$ is a closed subsapce

Question

Show $L^{\bot}$ and $H/L$(with $||\cdot||_{H/L}$) where $H$ is hilbert and $L$ is a closed subsapce are isometric, given $||\cdot||_H$ is defined by the inner product. By now I have considered $T:L^{\bot}\to {H/L}$ by $T(x)=x+L$. My problem is that I can't see how I derive anything from the fact that the norm is defined by the inner product and how I can compare inner products of two different spaces. I would appreciate your help.

Answer 1

Well this isn't actually that hard. Start with the following: $L^\perp$ is an orthogonal complement so that you can write every $x\in H$ uniquely as $x=y+y'$ with $y\in L, y'\in L^\perp$. Thus $x+L=y'+L$. This already proves that $T$ is a (even linear) bijection $L^\perp \to H/L$. What we need is that it is an isometry.

Now remember that the norm of $H/L$ is defined by $\|x+L\|_{H/L} := \inf \{ \|z\|_H \mid z\in x+L\}$. Therefore it we need to show that $\|x+L\|_{H/L} = \|y'\|_H$ in terms of the orthogonal decomposition from above. Clearly $\|y'\|_H \geq \|x+L\|_{H/L}$ just by definition. The other inequality is where the inner product comes in: Every $z\in x+L$ can be decomposed as $z=w+w'$ with $w\in L$, $w'\in L^\perp$. In fact $w'=y'$ because $0\equiv z-x = (w+w')-(y+y') \equiv w'-y' \mod L$ and thus $w'-y'\in L^\perp\cap L=0$. Thus we compute $\|z\|_H = \sqrt{\|w\|_H^2+\|w'\|_H^2} = \sqrt{\|w\|_H^2+\|y'\|_H^2} \geq \|y'|_H$. Taking the infimum over all $z\in x+L$ we find $\|x+L\|_{H/L} \geq \|y'\|_H$ as desired.