1
$\begingroup$

Consider a sequence of functions $f_n:[0,1] \rightarrow (0,1)$ and a function $g:[0,1] \rightarrow (0,1) $ such that

\begin{equation} \forall x,y, \lim_{n \rightarrow +\infty}{\frac{f_n(x)}{f_n(y)}}=\frac{g (x)}{g(y)} \end{equation}

and \begin{equation} \forall n \in \mathbb {N}, \int{f_n(x)dx=1} \end{equation}

I am trying to show that these conditions imply \begin{equation} \forall x \in [0,1], f_n (x) \rightarrow \frac{g(x)}{\int{g(y)dy}} \end{equation}

Any help would be much appreciated. Thanks!

$\endgroup$
2
$\begingroup$

If we also have that for a particular $x \in [0,1],\inf_n(x)f_n(x) = k_x>0,$ then we have that $$\left|\frac{f_n(y)}{f_n(x)}\right| \leq h(y) := \frac{1}{k_x} <\infty .$$ Now, since $h(y)$ is integrable over $[0,1]$, by dominated convergence we have that $$\lim_{n \to \infty}\int \frac{f_n(y)}{f_n(x)}dy = \int \lim_{n \to \infty}\frac{f_n(y)}{f_n(x)}dy.$$

Now, we have that $$\lim_{n \to \infty}\frac{1}{f_n(x)} = \lim_{n \to \infty} \left(\frac{\int f_n(y)dy}{f_n(x)}\right) = \lim_{n \to \infty}\left( \int \frac{f_n(y)}{f_n(x)}dy \right) = \int \lim_{n \to \infty}\frac{f_n(y)}{f_n(x)}dy = \int \frac{g(y)}{g(x)}dy = \frac{\int g(y)dy}{g(x)}.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! One question: I don't understand how you define $h$. Shouldn't $h$ be independent of $n$ to apply the dominated convergence theorem? $\endgroup$ – Oliv Mar 12 '17 at 11:23
  • $\begingroup$ @Oliv ... yes, i left out an important detail. Define it as the supremum over $n$....by your second condition in the question it is integrable for all $n$ so it is still integrable. $\endgroup$ – David Mar 12 '17 at 12:24
  • $\begingroup$ @oliv actually I'm not sure if that's true...let me think about it. $\endgroup$ – David Mar 12 '17 at 12:30
  • $\begingroup$ Actually we don't even know that $h$ is well defined, do we? I am wondering if we need extra conditions, for instance that the sequence $f_n(x)$ has a positive lower bound for all $x$, for the result to be true. Or is that implied by the assumptions? $\endgroup$ – Oliv Mar 12 '17 at 12:38
  • $\begingroup$ I had originally thought it was implied by the assumptions, but that seems unfounded. Are you integrating over a finite set, or are you integrating over a subset of the reals with infinite (lebesque) measure? If finite, then the condition $\inf_n f_n(x)=m>0$ is sufficient. $\endgroup$ – David Mar 12 '17 at 12:49

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.