TheoremBase

Proof

Fix x,y∈Rnx,y\in\mathbb{R}^{n} and t∈Rt\in\mathbb{R}. Throughout, zβ‹…zβ€²z\cdot z' is the dot product; by claim 1 of Properties of the Euclidean Norm on Rn\mathbb{R}^n: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square one has βˆ₯zβˆ₯2=zβ‹…z\lVert z\rVert^{2}=z\cdot z for every z∈Rnz\in\mathbb{R}^{n}, and the symmetry and bilinearity of the dot product used below are claims 1--5 of The Dot Product on Rn\mathbb{R}^n Is Symmetric, Bilinear and Positive Definite. Arithmetic with real numbers uses the axioms of the field of real numbers; we write 22 for 1+11+1, so that s+s=2ss+s=2s for every real ss.

Step 1 (expansion of the left-hand side). Put z=t x+(1βˆ’t) yz=t\,x+(1-t)\,y. Additivity and homogeneity in the first argument (claims 2 and 4) give

zβ‹…z=t (xβ‹…z)+(1βˆ’t) (yβ‹…z),z\cdot z=t\,(x\cdot z)+(1-t)\,(y\cdot z),

and the corresponding statements in the second argument (claim 5) give xβ‹…z=t (xβ‹…x)+(1βˆ’t) (xβ‹…y)x\cdot z=t\,(x\cdot x)+(1-t)\,(x\cdot y) and yβ‹…z=t (yβ‹…x)+(1βˆ’t) (yβ‹…y)y\cdot z=t\,(y\cdot x)+(1-t)\,(y\cdot y). Since yβ‹…x=xβ‹…yy\cdot x=x\cdot y by claim 1, substituting and collecting terms yields

βˆ₯zβˆ₯2=t2 βˆ₯xβˆ₯2+2 t(1βˆ’t) (xβ‹…y)+(1βˆ’t)2 βˆ₯yβˆ₯2.\lVert z\rVert^{2}=t^{2}\,\lVert x\rVert^{2}+2\,t(1-t)\,(x\cdot y)+(1-t)^{2}\,\lVert y\rVert^{2}.

Step 2 (expansion of the squared distance). By claim 3 (differences in the first argument) and claim 5 (differences in the second argument), together with claim 1,

βˆ₯xβˆ’yβˆ₯2=(xβˆ’y)β‹…(xβˆ’y)=βˆ₯xβˆ₯2βˆ’2 (xβ‹…y)+βˆ₯yβˆ₯2.\lVert x-y\rVert^{2}=(x-y)\cdot(x-y)=\lVert x\rVert^{2}-2\,(x\cdot y)+\lVert y\rVert^{2}.

Step 3 (comparison). Substituting the identity of Step 2 into the right-hand side of the assertion and expanding,

t βˆ₯xβˆ₯2+(1βˆ’t) βˆ₯yβˆ₯2βˆ’t(1βˆ’t) βˆ₯xβˆ’yβˆ₯2=(tβˆ’t(1βˆ’t))βˆ₯xβˆ₯2+((1βˆ’t)βˆ’t(1βˆ’t))βˆ₯yβˆ₯2+2 t(1βˆ’t) (xβ‹…y).t\,\lVert x\rVert^{2}+(1-t)\,\lVert y\rVert^{2}-t(1-t)\,\lVert x-y\rVert^{2} =\bigl(t-t(1-t)\bigr)\lVert x\rVert^{2}+\bigl((1-t)-t(1-t)\bigr)\lVert y\rVert^{2}+2\,t(1-t)\,(x\cdot y).

Now tβˆ’t(1βˆ’t)=tβˆ’t+t2=t2t-t(1-t)=t-t+t^{2}=t^{2} and (1βˆ’t)βˆ’t(1βˆ’t)=(1βˆ’t)(1βˆ’t)=(1βˆ’t)2(1-t)-t(1-t)=(1-t)(1-t)=(1-t)^{2}, so the right-hand side coincides with the expression obtained in Step 1, namely βˆ₯t x+(1βˆ’t) yβˆ₯2\lVert t\,x+(1-t)\,y\rVert^{2}. Since xx, yy and tt were arbitrary, the identity holds in general.

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