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!


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

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  • $\begingroup$ 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

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