Prove that the set of those $c$ is countable

  1. The **mean values theorem** says that there exists a $c∈(u,v)$ such that $$f(v)-f(u)=f′(c)(v-u)$$ My question is: Assume that $u$ is a root of $f$, hence we obtain $$f(v)=f′(c)(v-u)$$ Assume that $f$ is a non-zero analytic function in the whole real line. My interest is about the real $c∈(u,v)$. I believe that the set of those $c$ is contable, otherwise we conclud that $f′(c)$ is constant (in an open set containing $c$) since $v$ is a constant and hence $f$ is identically zero. Also, I think that the set of those $c$ is countable (for analytic functions) but I am not able to prove that.
     
  2. jcsd
  3. For all c that satisfy that equation, [itex] f'(c)(v-u)-f(v)=0 [/itex]. The LHS is an analytic function in c (v and u are constant) and is not identically zero unless f is linear. That any non-zero analytic function has countable zeroes is well known. (http://planetmath.org/ZeroesOfAnalyticFunctionsAreIsolated is a proof for complex functions, but the idea is the same)

    Edit: My link only proves that the zeroes are isolated. Here is how you would show that the set of zeroes in countable: Let a be a zero. If all other zeroes are separated from a by at least ε, let q be a rational number which is no more than ε/3 away from a. Each rational number can be associated with a zero in this way. The cardinality of the set of zeroes is the same as for this set of rational numbers and rational numbers are countable.
     
    Last edited: Apr 27, 2013
  4. Fredrik

    Fredrik 10,298
    Staff Emeritus
    Science Advisor
    Gold Member

    Type ## instead of $.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted