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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?

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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 $c$.

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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 $$ \|a-c\|_{\infty} = \|a-b\|_{\infty}+\|c-b\|_{\infty} $$

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  • $\begingroup$ 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. $\endgroup$ – Amir Hassan Jun 22 '20 at 21:49
  • $\begingroup$ @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. $\endgroup$ – Reinhard Meier Jun 23 '20 at 7:37

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