Inner Product Space Inequality

Question

Let $X$ be an inner product space, let $L\subset X$ be a vector subspace, let $x\in X$, let $d=\inf_{y\in L}\left\|x-y\right\|$, and let $y_1,y_2\in L$.

Could someone please provide a hint of how to show that

$$\left\|y_1-y_2\right\|\leq\sqrt{\left\|y_1-x\right\|^2-d^2}+\sqrt{\left\|y_2-x\right\|^2-d^2}\tag*{?}$$

I have tried the parallelogram law and orthogonal decomposition to no avail. The case where $x\in\overline L$ is straightforward.

Answer 1

Let $M \subset L$ be the linear subspace spanned by $y_1, y_2$, and $y^\ast$ the orthogonal projection of $x$ onto $M$. Then for $i = 1,2$:

• $\langle y_i - y^\ast, x - y^\ast\rangle = 0$, and therefore the Pythagorean theorem
• $\Vert y_i - x \Vert^2 = \Vert y_i - y^\ast \Vert^2 + \Vert x - y^\ast \Vert^2 $ holds.
• $d=\inf_{y\in L}\left\|x-y\right\| \le \Vert x - y^\ast \Vert $ .

Now use that to estimate $$ \Vert y_1 - y_2 \Vert \le \Vert y_1 - y^\ast \Vert + \Vert y_2 - y^\ast \Vert $$