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Where does the following equality comes from? \begin{equation} 2\|z_n-x\|^2+2\|z_{n+1}-x\|^2-4\|y_n-x\|^2=\|z_{n+1}-z_{n}\|^2 \end{equation} where $y_n := (z_n+z_{n+1})/2$, and $z_n$, $z_{n+1}$, $x$ are elements of an inner product vector space, being $\|\cdot\|$ the norm induced by the inner product.

It is trivial to prove the equality from left to right, but I can't figure out how to go the other way around 'intuitively'.

By 'intuitively' I mean, I have the square of a distance between two points and I want to write this square distance in terms of the square distance between each of these points and a third one.

Any idea?

Some context

I came across with this equality while reading the proof of proposition 2, chapter 20, of the book "A Modern Approach to Probability Theory" by Fristedt and Gray. The aforementioned proposition proves the existence and uniqueness of the orthogonal projection in a Hilbert space.

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    $\begingroup$ Not verified, but worth a try... Let $p$ be the projection of $x$ on the line $z_n,z_{n+1}$. Apply Pythagorean theorem on the triangles having $x,p$ as one side. $\endgroup$ Jan 23, 2022 at 10:12
  • $\begingroup$ Thanks for the idea! But it doesn't seem to work. $\endgroup$ Jan 23, 2022 at 10:42
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    $\begingroup$ This looks like the parallelogram law for general quadrilaterals. $\endgroup$
    – Phicar
    Jan 23, 2022 at 10:58
  • $\begingroup$ Thank you, @Phicar, your hint did the trick! $\endgroup$ Jan 23, 2022 at 12:43
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    $\begingroup$ If you familiar with euclidean geometry, then you can compare it to the formula for the length of median of a triangle en.m.wikipedia.org/wiki/Median_(geometry) $\endgroup$
    – richrow
    Jan 23, 2022 at 13:00

2 Answers 2

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As Phicar pointed out, this looks like the parallelogram law.

The parallelogram law

Indeed, let $x$ and $y$ be elements of an inner product vector space, and let $\|\cdot\|$ be the norm induced by the inner product. Then, straightforward calculations give us the following equations: \begin{equation} \|x+y\|^2=\|x\|^2+\|y\|^2+2\langle x,y\rangle \end{equation} and \begin{equation} \|x-y\|^2=\|x\|^2+\|y\|^2-2\langle x,y\rangle\text{, } \end{equation} where $\langle\cdot,\cdot\rangle$ is the inner product. Summing up this two equalities we obtain the so called parallelogram law, that is \begin{equation} \|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2\text{. } \end{equation}

Application of the parallelogram law in the current context

With all of this in mind, we can rearange the parallelogram law and get \begin{equation} \|x-y\|^2=2\|x\|^2+2\|y\|^2-\|x+y\|^2\text{. } \end{equation}

Defining $x:=z_n-u$ and $y:= z_{n+1}-u$ we get that \begin{equation} \|z_n-z_{n+1}\|^2=2\|z_n-u\|^2+2\|z_{n+1}-u\|-\|z_{n}+z_{n+1}-2u\|^2\text{, } \end{equation} hence, getting the $2$ out of the norm of the last term we get that \begin{equation} \|z_n-z_{n+1}\|^2=2\|z_n-u\|^2+2\|z_{n+1}-u\|-4\left\|\frac{z_{n}+z_{n+1}}{2}-u\right\|^2\text{, } \end{equation} and this equation is just the result we were seeking.

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    $\begingroup$ Nice! +1. Geometrically, I guess, is placing one of the vertices of the quadrilateral as the midpoint of a side. $\endgroup$
    – Phicar
    Jan 23, 2022 at 12:59
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You can write using the parallelogram viewpoint $$ \vec{XZ}_{n} + \vec{XZ}_{n+1}=2 \vec{XY_n}$$ and take the (squared) norm of both sides $$ \mathbf{d}^T \mathbf{d} =4 \|\mathbf{y}_n-\mathbf{x}\|^2 $$ where $ \mathbf{d}=(\mathbf{z}_n-\mathbf{x})+(\mathbf{z}_{n+1}-\mathbf{x})$.

Expanding the left term $$ \|\mathbf{z}_n-\mathbf{x}\|^2+ \|\mathbf{z}_{n+1}-\mathbf{x}\|^2+ (\mathbf{z}_n -\mathbf{x})^T (\mathbf{z}_{n+1}\color{#FF0000}{-\mathbf{z}_{n}+\mathbf{z}_{n}}-\mathbf{x})+ (\mathbf{z}_{n+1} -\mathbf{x})^T (\mathbf{z}_{n}\color{#FF0000}{-\mathbf{z}_{n+1}+\mathbf{z}_{n+1}}-\mathbf{x}) $$

the last two terms become using the trick in red $$ \|\mathbf{z}_n-\mathbf{x}\|^2+ \|\mathbf{z}_{n+1}-\mathbf{x}\|^2+ (\mathbf{z}_n -\mathbf{x})^T (\mathbf{z}_{n+1}-\mathbf{z}_{n})+ (\mathbf{z}_{n+1} -\mathbf{x})^T (\mathbf{z}_{n}-\mathbf{z}_{n+1}) $$ Factorizing the two last terms give $$ \left[ (\mathbf{z}_n -\mathbf{x}) - (\mathbf{z}_{n+1} -\mathbf{x}) \right]^T (\mathbf{z}_{n+1}-\mathbf{z}_{n}) = - \| \mathbf{z}_{n+1}-\mathbf{z}_{n} \|^2 $$ Putting alltogether gives the expected result

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  • $\begingroup$ Thanks! Nice proof as well! $\endgroup$ Jan 24, 2022 at 12:42

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