# Prove that the given linear map is injective

Let $n \in \mathbb{N}$ be a fixed positive integers and $B \in \mathbb{R}_+$ also be fixed. For a fixed $M>0$, let $f:[-B,B]^n \to \mathbb{R}$ be given by $f(x_1,\ldots,x_n)=\sum_{i=1}^n x_i M^i$. My aim is to prove the following:

Prove that one can always find a fixed positive real number $M$ (dependent only on $B$ and $n$) such that $f$ is an injective map.

If such an $M$ does not exist in the above question, one can assume that each $x_i$ can take only a fixed set of countable values in $[-B,B]$. In other words, the map $f$ is defined only for a countable subset of $[-B,B]^n$. In this reduced case, can $f$ be injective?

My attempt: Let $f(x_1,\ldots,x_n)=f(y_1,\ldots,y_n)$ for some $(x_1,\ldots,x_n),(y_1,\ldots,y_n) \in [-B,B]^n$. Thus, $\sum_{i=1}^n (x_i-y_i)M^i=0$. After this step, I am stuck and unable to procced further about how to choose such an $M$ such that the above equation does't have any solution.

• Is it given that $a_i \neq b_i$ for all $1 \leq i \leq n$? – Noble Mushtak Mar 28 '16 at 22:50
• Yeah.Or atleast one $a_i \neq b_i$. – pikachuchameleon Mar 28 '16 at 22:56
• Are all the $a's$ and $b's$ specified in advance? Then you just have a polynomial with some non-zero coefficients. – lulu Mar 28 '16 at 23:07
• @lulu: I edited the question a bit. Please check. – pikachuchameleon Mar 29 '16 at 18:16
• This doesn't work. If $M>1$ then take any $b_1\neq 0$ in the range and define $a_n=\frac {b_1}{M^{n-1}}$, all others $0$. If $M≤1$ then take $b_2\neq 0$ in the range and let $a_1=b_2\,M$, all others $0$. – lulu Mar 29 '16 at 18:25

The general statement is definitely false for $n > 1$. $f(0, 0)=0$, but we can also choose some $0 < \epsilon < \min(B, BM)$ so that $\left(-\epsilon, \frac \epsilon M\right) \in [-B, B]^2$ and thus in the domain of $f$. Thus, we have the following: $$f\left(-\epsilon, \frac \epsilon M\right)=-\epsilon M+\frac \epsilon M M^2=-\epsilon M+\epsilon M=0$$ Therefore, $f(0, 0)=f\left(-\epsilon, \frac \epsilon M\right)$ and $f$ is not injective.

However, I think the reduced statement is actually true. Let $D$ be the domain of $f$. For any two unequal ${\bf x}, {\bf y} \in D$, $f({\bf x})=f({\bf y})$ means that $\sum_{i=1}^n ({\bf x}_i-{\bf y}_i)M^i=0$. Since the left side of this equation is a non-zero polynomial of degree at most $n$ in terms of $M$, there are at most $n$ values of $M$ which satisfies this equality.

We have the following:

• Since $D$ is countable, there are a countable number of pairs of unequal ${\bf x}, {\bf y} \in D$.
• Given a specific pair of ${\bf x}, {\bf y} \in D$, there are a finite number of $M$ where $f({\bf x})=f({\bf y})$.

Therefore, overall, there are a countable set of $M$ where $f({\bf x})=f({\bf y})$ is true for any ${\bf x}, {\bf y} \in D$. On the other hand, there are an uncountable number of positive real numbers, so choose a positive number that is not in the described countable set. For this $M$, $f({\bf x}) \neq f({\bf y})$ for all ${\bf x}, {\bf y} \in D$, which means $f$ is injective.

• I modified the question a bit. Please verify. – pikachuchameleon Mar 29 '16 at 18:16
• @pikachuchameleon I have revised my answer. Does it answer your question? – Noble Mushtak Mar 29 '16 at 20:23
• Yeah, more or less that resolves the question. But, how do you make sure that $\frac{\epsilon}{M} \leq B$ in the counter-example? – pikachuchameleon Mar 29 '16 at 22:35
• $\epsilon$ can be any positive number, so we simply pick a $0 < \epsilon < BM$. If you want a specific example, $\epsilon=\frac{BM} 2$ works. – Noble Mushtak Mar 30 '16 at 0:15
• Okay. Then how do you ensure that $BM/2 < B$. Remember that I want injectivity on the domain $[-B,B]$. Also, in the second part of the above solution regarding the reduced case, it doesn't matter if they are bounded right? I think your solution doesn't use that. – pikachuchameleon Mar 30 '16 at 14:05