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If $u$ and $v$ are functions on a function space, what is an easy way to see that Pythagorean holds for functions as it does for triangles:

Suppose u and v are orthogonal functions in V. Then $\|u+v\|^{2}=\|u\|^{2}+\|v\|^{2}$

This is used later to prove for minimization criteria for orthogonal spaces:

Suppose U is a finite-dimensional subspace of V. Then two vectors u and v.

Then $\left\|v-P_{U} v\right\| \leq\|v-u\|$

Furthermore, the inequality above is an equality if and only if $u=P_{U} v$

where $P_{U} v$ is the orthogonal projection of $v$ onto subspace of U

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  • $\begingroup$ Use the inner product definition for that function space and the fact that $\|u\|^2=\langle u,u \rangle$. $\endgroup$ – Anurag A Jul 17 '19 at 18:13
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    $\begingroup$ @Azif00 the question is in the first line: "what is an easy…" $\endgroup$ – Anurag A Jul 17 '19 at 18:16
  • $\begingroup$ I wouldn't say it holds for triangles nor for functions but rather it holds for inner product spaces, making the question moot. $\endgroup$ – rschwieb Jul 17 '19 at 18:29
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In general, for any orthogonal vectors $x$ and $y$ in a inner product space $( \textsf V , \langle \cdot , \cdot \rangle )$ the equality $$\| x+y \|^2 = \| x \|^2 + \| y \|^2$$ is still true and the proof is the following :

$$\begin{align} \| x+y \|^2 &= \langle x+y,x+y \rangle \\ &= \langle x,x+y \rangle + \langle y,x+y \rangle \\ &= \langle x,x \rangle + \langle x,y \rangle + \langle y,x \rangle +\langle y,y \rangle \\ &= \langle x,x \rangle + \langle y,y \rangle \\ &= \| x \|^2 + \| y \|^2 \end{align}$$ Note : $\langle x,y \rangle = \langle y,x \rangle =0$ by definition since $x$ and $y$ are orthogonal.

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By linearity ad distributivity (two properties of any linear vector space) we have $$ ||u+v||^2 \equiv \left< (u+v), (u+v) \right> = \left< u, u \right> + \left< u, v \right> + \left< v,u \right> + \left< v, v \right> =||u||^2 +||v||^2 + \left< u, v \right> + \left< v,u \right> $$ and since $u$ and $v$ are orthogonal, by definition the last two terms are zero.

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