0
$\begingroup$

Suppose we are given that $$\sum_{i =a} ^b f (i)g (i) =\sum_{i =a} ^b f (i)h (i) \tag {1} $$ when can we conclude that $$\forall i =a,…,b:g(i)=h(i) \tag {2} $$ ?

For example, if $g, h $ are both linear and $\sum_i i f(i)\not = 0$ then $(1)\implies (2)$.

Is there a name for this theorem or set of functions for which $(1)\implies (2)$?

Let's suppose $f,g,h:\mathbb R\to\mathbb R$.

$\endgroup$
  • 1
    $\begingroup$ You might also want to assume that $\sum\limits_iif(i)\ne0$ for the implication to hold. A name? I doubt there is one. $\endgroup$ – Did Apr 17 '15 at 11:54
  • $\begingroup$ You should be more specific on the involved functions; what is their domain and codomain? What means "linear" in this context? In general the conclusion is false: if $f(1)=f(2)=1$, $g(1)=1$, $g(2)=-1$, $h(1)=h(2)=0$ then also $f(1)g(1)+f(2)g(2)=f(1)h(1)+f(2)h(2)=0$, but $g\ne h$. $\endgroup$ – user 59363 Apr 17 '15 at 12:24
  • $\begingroup$ @user59363: I meant this definition of linear, which your example $g$ doesn't meet. $\endgroup$ – Xodarap Apr 17 '15 at 23:09
  • 1
    $\begingroup$ Of course is my $g$ not a linear function and neither are my $f$ and $h$: their domain is the set $\{1,2\}$. In order to have linear functions their domain must be a vector space. Therefore I thought you should be more specific about domain and codomain of your functions. As the problem is stated now, one has to assume the domain of your functions $f,g,h$ is the finite set of integers $\{a,a+1,\dots, b\}$. However, it makes no sense to speak about linearity of such functions. $\endgroup$ – user 59363 Apr 18 '15 at 18:09
  • $\begingroup$ @user59363: sure, added "Let's suppose $f,g,h:\mathbb R\to\mathbb R$." $\endgroup$ – Xodarap Apr 19 '15 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.