# Inner product space proof of equality $\|a-c\|=\|a-b\|+\|c-b\|$

is anybody out there who can help with this little problem? I know the definition of inner product space, but I don't know how to do the proof of the exercise below.

Let $$a,b,c$$ be elements of an inner product space. Show that the following equality $$\|a-c\|=\|a-b\|+\|c-b\|$$ holds iff there exists a real number $$\mu\in[0,1]$$ s.t. $$b=\mu a+(1-\mu)c$$.

Can you give an example that shows that this is not true in an arbitrary normed space?

In general, if $$\|\cdot\|$$ is induced by an inner product, then $$\|x+y\|=\|x\|+\|y\|$$ if and only if one of $$x$$ and $$y$$ is a nonnegative scalar multiple of the other.

Now let $$x=a-b$$ and $$y=b-c$$. The given condition implies that $$\|x+y\|=\|x\|+\|y\|$$. Therefore, one of $$x$$ and $$y$$ is a nonnegative scalar multiple of the other. Without loss of generality, let $$y=kx$$ for some $$k\ge0$$. Then $$b-c=k(a-b)$$. Hence $$b=\frac{k}{k+1}a+\frac{1}{k+1}c$$ and you may take $$\mu=\frac{k}{k+1}$$.

The case is different if the norm is not induced by an inner product. E.g. let $$a=(1,0),\,b=(0,0)$$ and $$c=(0,1)$$ in $$\mathbb R^2$$. We have $$\|a-c\|_1=2=\|a-b\|_1+\|c-b\|_1$$, but $$b$$ does not lie on the line segment delimited by $$a$$ and $$c$$.

If $$b=\mu a+(1-\mu)c$$ and $$\mu\in[0,1],$$ then $$\|a-b\|+\|c-b\| \\ = \|a-\mu a-(1-\mu)c\|+\|c-\mu a-(1-\mu)c\| \\ = \|(1-\mu)(a-c)\|+\|(-\mu)(a-c)\| \\ = (1-\mu)\,\|a-c\| + \mu\,\|a-c\| \\ = \|a-c\|$$ If $$\|a-c\|=\|a-b\|+\|c-b\|$$ and the norm is derived from an inner product, we get the following: $$\begin{eqnarray} \|a-c\| & = & \|a-b\|+\|c-b\| \\ \|a-c\|^2 & = & \|a-b\|^2+\|c-b\|^2 \\ & & +2\cdot\|a-b\|\cdot\|c-b\| \\ \langle a-c,a-c\rangle & = & \langle a-b,a-b\rangle+\langle c-b,c-b\rangle \\ & & +2\cdot\|a-b\|\cdot\|c-b\| \\ \langle a,a\rangle-\langle c,a\rangle-\langle a,c\rangle+\langle c,c\rangle & = & \langle a,a\rangle-\langle b,a\rangle-\langle a,b\rangle+\langle b,b\rangle \\ & & +\langle c,c\rangle-\langle b,c\rangle-\langle c,b\rangle+\langle b,b\rangle \\ & & +2\cdot\|a-b\|\cdot\|c-b\| \\ -\langle c,a\rangle+\langle b,a\rangle+\langle c,b\rangle-\langle b,b\rangle & & \\ -\langle a,c\rangle+\langle a,b\rangle+\langle b,c\rangle-\langle b,b\rangle & = & 2\cdot\|a-b\|\cdot\|c-b\| \\ \langle c-b ,b-a\rangle +\langle b-a,c-b\rangle & = & 2\cdot\|a-b\|\cdot\|c-b\| \\ 2\operatorname{Re}\; \langle c-b ,b-a \rangle & = & 2\cdot\|a-b\|\cdot\|c-b\| \\ \operatorname{Re}\; \langle c-b ,b-a \rangle & = & \|c-b\|\cdot\|b-a\| \\ \end{eqnarray}$$ From $$\operatorname{Re}(z)\leq |z|$$ and from the Cauchy-Schwarz inequality we conclude $$\operatorname{Re}\; \langle c-b ,b-a \rangle \leq |\,\langle c-b ,b-a \rangle\,| \leq \|c-b\|\cdot\|b-a\|$$ As we have shown $$\operatorname{Re}\; \langle c-b ,b-a \rangle=\|c-b\|\cdot\|b-a\|$$, this means $$|\,\langle c-b ,b-a \rangle\,| = \|c-b\|\cdot\|b-a\|$$ The Cauchy-Schwarz inequality tells us that this is a sufficient condition for $$c-b$$ and $$b-a$$ being linearly dependent, i.e. there are numbers $$\lambda_1$$ and $$\lambda_2$$ with $$(\lambda_1,\lambda_2)\neq(0,0)$$ such that $$\lambda_1(c-b) + \lambda_2 (b-a) =0.$$ In the following, we only consider the non-degenerate case $$a\neq c.$$ Then $$\lambda_1\neq \lambda_2$$. (If we had $$\lambda_1=\lambda_2$$, then $$\lambda_1(c-b) + \lambda_2 (b-a) =0$$ would imply $$a=c$$.)

From $$\lambda_1(c-b) + \lambda_2 (b-a) =0$$ we get $$b=\frac{\lambda_2}{\lambda_2-\lambda_1}\cdot a +\frac{-\lambda_1}{\lambda_2-\lambda_1}\cdot c$$ Now we can set $$\mu = \frac{\lambda_2}{\lambda_2-\lambda_1}$$ and we have $$b=\mu a +(1-\mu) c$$ In order to show that this $$\mu$$ must be in $$[0,1],$$ we can put the $$b$$ into $$\|a-c\|=\|a-b\|+\|c-b\|$$, like in the first part of the proof, and we get $$\|a-c\|=\left(|1-\mu| +|\mu|\right)\cdot \|a-c\|$$ which means $$|1-\mu| +|\mu| = 1$$ In other words, the sum of the distances of the complex number $$\mu$$ to the points $$0$$ and $$1$$ in the complex plane must be $$1.$$ This can obviously be fulfilled if and only if $$\mu\in[0,1].$$

Let $$a=\binom{1}{0},\;\;b=\binom{0}{1},\;\;c=\binom{-1}{0}.$$ Then there is no $$\mu$$ such that $$b=\mu a+(1-\mu)c$$, but $$\|a-c\|_{\infty} = \|a-b\|_{\infty}+\|c-b\|_{\infty}$$

• Dear Reinhard, can you please help me to write $||a-c||=||a-b||+||c-b||$ out? I've tried but failed several times, even my friends tried too. I hope you can give a hint or two. – Amir Hassan Jun 22 '20 at 21:49
• @AmirHassan I am sorry, I omitted complex numbers and the possibility of $\langle u,v\rangle\neq \langle v,u\rangle$ in my first attempt. I have now added the details and made the proof valid in general. – Reinhard Meier Jun 23 '20 at 7:37