# Bounding the dot product of two planar unit vectors.

Does there exist a continuous, monotone increasing function $$f\colon[0,2]\to [0,1]$$, satisfying $$f(0)=0$$ and $$f(1)=1$$, such that for all vectors $$(a_1,b_1),(a_2,b_2)\in \mathbb{R}^2$$ of unit length, i.e. $$a_1^2+b_1^2= a_2^2+b_2^2 =1$$, one has

$$|(a_1,b_1)\cdot (a_2,b_2)|^2=|a_1a_2+b_1b_2|^2 \leq f(a_1^2+b_2^2)$$

If so, can one obtain a simple formula for such a function? Is there an obvious candidate?

Some motivation: If $$a_1^2+b_2^2=0$$, then certainly $$|(a_1,b_1)\cdot (a_2,b_2)|^2=0$$. On the other hand, by Cauchy-Schwarz, we have $$|(a_1,b_1)\cdot (a_2,b_2)|^2 \leq |(a_1,b_1)|^2 |(a_2,b_2)|^2 =1$$ with equality if and only if $$(a_1,b_1)=\lambda(a_2,b_2)$$ for $$\lambda=\pm1$$, which implies $$a_1^2+b_2^2=1$$.

EDIT: On a previous posting of this question I supposed the domain of $$f$$ should be [0,1], which can be seen to be ill posed.

• If $S^1$ denotes the unit circle in $\Bbb{R}^2$, the range of the map $$S^1\times S^1\ \longrightarrow\ \Bbb{R}:\ ((a_1,b_1),(a_2,b_2))\ \longmapsto\ a_1^2+b_2^2,$$ is the interval $[0,2]$. But your function $f$ has $[0,1]$ as its domain; what happens when $a_1^2+b_2^2>1$? – Servaes Feb 7 at 21:05
• if $(a_1,b_1), (a_2,b_2) = (1,0),(0,1)$ respectively then $(a_1^2 + b_2^2) = 2$ which is outside the domain of $f.$ – Doug M Feb 7 at 21:06
• Thanks @Servaes – Aerinmund Fagelson Feb 7 at 22:48
• and also @Doug_M. Excellent points! I will edit my question accordingly. Many thanks :) – Aerinmund Fagelson Feb 7 at 22:49
• To be clear; you require that $f(x)=1$ for all $x\in[1,2]$? – Servaes Feb 7 at 22:53

Disclaimer: The answer below is horrible; there is clearly a nice geometric argument somewhere, but it's rather late here and I don't see it. Perhaps tomorrow I will find a nicer answer.

The short answer is $$f(x)=x(2-x)$$.

Let $$S^1\subset\Bbb{R}^2$$ denote the unit circle, and for a unit vector $$u\in S^1$$ let $$u_1$$ and $$u_2$$ denote its first and second coordinate, so that $$u=(u_1,u_2)$$. Then for every $$c\in[0,2]$$ we have $$f(c)\geq\max_{\substack{u,v\in S_1\\u_1^2+v_2^2=c}}|u\cdot v|^2 =\max_{\substack{u,v\in S_1\\u_1^2+v_2^2=c}}\left(u_1^2v_1^2+2u_1u_2v_1v_2+u_2^2v_2^2\right).$$ For all $$u,v\in S^1$$ corresponding to a given $$c\in[0,2]$$ we have $$u_1^2+v_2^2=c$$. Of course $$u_1^2+u_2^2=v_1^2+v_2^2=1$$ and so $$v_1^2=1-(c-u_1^2)$$, so we can rewrite the above as $$\begin{eqnarray*} u_1^2v_1^2+2u_1u_2v_1v_2+u_2^2v_2^2 &=&u_1^2(1-(c-u_1^2))+2u_1u_2v_1v_2+(1-u_1^2)(c-u_1^2)\\ &=&u_1^2(1-(c-u_1^2))\pm2u_1\sqrt{1-u_1^2}\sqrt{1-(c-u_1^2)}\sqrt{c-u_1^2}+(1-u_1^2)(c-u_1^2),\\ &=&2u_1^4-2cu_1^2+c+2u_1\sqrt{(u_1^2-1)(u_1^2-c)(u_1^2-(c-1))}, \end{eqnarray*}$$find where the last equality holds after changing the sign of $$u_1$$ by replacing $$u$$ and $$v$$ by $$-u$$ and $$-v$$ if necessary; this does not change the value of $$|u\cdot v|^2$$. The above is a function of $$u_1$$, which we denote by $$g_c$$. We determine the maximum of $$g_c(x)$$ with the constraint that $$x^2\in[0,1]$$ and $$c-x^2\in[0,1]$$, and $$x\geq0$$. We have $$g_c(x)=2x^4-2cx^2+c+2x\sqrt{(x^2-1)(x^2-c)(x^2-(c-1))}.$$ The boundary points are either $$0$$ and $$\sqrt{c}$$, if $$c\leq1$$, or $$\sqrt{c-1}$$ and $$1$$, if $$c\geq1$$, and we have $$g_c(0)=g_c(\sqrt{c})=c,\qquad g_c(\sqrt{c-1})=g_c(1)=2-c.$$ For the maxima in the interior we compute the derivative $$\begin{eqnarray*} \frac{dg_c}{dx}(x) &=&8x^3-4cx+2\sqrt{(x^2-1)(x^2-c)(x^2-(c-1))} +\frac{2x^2(3x^4-4cx^2+(c^2+c-1))}{\sqrt{(x^2-1)(x^2-c)(x^2-(c-1))}}\\ &=&4x(2x^2-c)+2\left(1+\frac{x^2(3x^4-4cx^2+(c^2+c-1))}{(x^2-1)(x^2-c)(x^2-(c-1))}\right)\sqrt{(x^2-1)(x^2-c)(x^2-(c-1))}\\ &=&4x(2x^2-c)+2\frac{4x^6-6cx^4+2(c^2+c-1)x^2+(c-c^2)}{\sqrt{(x^2-1)(x^2-c)(x^2-(c-1))}}, \end{eqnarray*}$$ and setting this equal to zero and rearranging a bit yields $$4x^2(c-2x^2)\sqrt{(x^2-1)(x^2-c)(x^2-(c-1))}=4x^6-6cx^4+2(c^2+c-1)x^2+(c-c^2),$$ and after squaring, both are sextic polynomials in $$x^2$$. Collecting terms, it turns out that almost everything cancels and we are miraculously left with the following quadratic in $$x^2$$: $$(c-1)^2(2x^2-c)^2=0,$$ so when $$c\neq1$$ the unique interior extremum is at $$x=\sqrt{\frac{c}{2}}$$, where $$g_c\left(\sqrt{\frac{c}{2}}\right)=c(2-c).$$ Comparing this to the values at the boundary points, clearly $$c(2-c)\geq c$$ if $$c\leq1$$ and $$c(2-c)\geq2-c$$ if $$c\geq1$$, hence the maximum equals $$c(2-c)$$ for all $$c$$. This shows that the sharpest upper bound is given by $$f(x)=x(2-x)$$, i.e. $$f(a_1^2+b_2^2)=(a_1^2+b_2^2)(2-a_1^2-b_2^2).$$

Update: I have a geometric argument, though I don't find it entirely satisfying. Perhaps someone can improve/complete it?

With the notation as above, for $$u,v\in S^1$$ let $$w:=(u_1,v_2)$$. Then $$||w||^2=u_1^2+v_2^2=c$$, so $$w$$ is a point on the circle of radius $$\sqrt{c}$$ centered at the origin. Conversely, to every point $$w=(u_1,v_2)$$ on this circle correspond four pairs of vector $$(u,v)$$, $$(u,v')$$, $$(u',v)$$ and $$(u',v')$$. The picture below clarifies the situation: As before, for each $$c\in[0,2]$$ we want that $$f(c)\geq\max_{\substack{u,v\in S_1\\u_1^2+v_2^2=c}}|u\cdot v|^2 =\max_{\substack{u,v\in S_1\\||w||^2=c}}|\cos\theta|^2,$$ where $$\theta$$ denotes the angle between $$u$$ and $$v$$. It is not hard to see from the picture above that out of the four pairs $$(u,v)$$, $$(u,v')$$, $$(u',v)$$ and $$(u',v')$$ the angle is maximal for $$(u,v)$$, the pair of vectors that are both not in the same quadrant as $$w$$.

Then $$|\cos\theta|$$ is maximal if and only if $$|\cos\psi|$$ is minimal, where $$\psi$$ is the angle made by the opposite pair $$(u',v')$$, because $$\theta+\psi=\pi$$. This angle is minimal precisely when the hypothenuse of the right-angle triangle $$\Delta u'wv'$$ is minimal. I find it visually clear that this is the case precisely when $$u'w=v'w$$, which is equivalent to $$u_1=v_2$$, which is equivalent to $$u_1=v_2=\sqrt{\tfrac{c}{2}}$$, though this does not immediately give me $$|\cos\theta^2|=c(2-c)$$...

I made a visualization of the problem in Geogebra, perhaps it gives some insight.

• What a heroic calculation!! Thank-you @Servaes!. I had wanted a monotone function, which is easy to obtain by setting $f(x) = x(2-x)$ for $x\in[0,1]$ and $f(x)=1$ for $x\in[1,2]$, but that is certainly crude and the sharp bound you have derived is beautiful. I would also like a geometric proof, and will try and find one myself. I'll wait a day before accepting your answer in case you do find one and would like to add it :) – Aerinmund Fagelson Feb 8 at 8:48
• I have found some improvements, though no nice geometric proof yet. I will update later today. – Servaes Feb 8 at 9:28
• I have made some minor corrections. My other idea ended up becoming just as big a mess as this one, so I'll leave it at this. – Servaes Feb 9 at 20:11