Consider the following problem:
Suppose $f:[0,1] \to \mathbb{R}$ is a differentiable function with $f'(0)<0$, $f(0)=1$ and for all $x\in(0,1]:0<f(x)<1$. Calculate $$\lim_{n\to\infty}\int_0^nf\left(\frac{x}{n}\right)^ndx.$$
My attempt so far:
Let $$f_n:x\mapsto\chi_{[0,n]}(x)\cdot f\left(\dfrac{x}{n}\right)^n,$$ where $\chi_{[0,n]}$ is the indicator function on $[0,n]$. We wish to do two things:
- Show that $f_n\to h$ pointwise for some integrable function $h:\mathbb{R}^+\to\mathbb{R}$ and
- show that $|f_n|\leq k$ for some positively integrable $k:[0,1]\to \mathbb{R}^+$ so that we can use the dominated convergence theorem to show that $$\lim_{n\to\infty}\int_{[0,\infty)}f_n(x) \,dx = \int_{[0,\infty)}h(x)dx.$$
To do this, we first construct the function $$g:[0,1]\to\mathbb{R}:y\mapsto\begin{cases}\dfrac{\ln f(y)}{y}& y\in(0,1],\\ f'(0) & y=0.\end{cases}$$
This is a continuous extention of $\ln f(y)/y$ since $$\lim_{y\to0}\dfrac{\ln f(y)}{y}=f'(0)$$
by l'Hôpital's rule. The idea is to show that $$f\left(\dfrac{x}{n}\right)^n\leq e^{f'(0)x}$$ for all $x\in[0,n]$. To do this we can use the function $g$ as constructed above. Indeed, taking the natural log of this inequality gives us
$$n\ln f\left(\dfrac{x}{n}\right)\leq f'(0)x$$
which is equivalent to $$g(y)\leq f'(0),$$ for all $y\in[0,1]$. The problem is that I have trouble showing that this last inequality is true. Although I'm convinced it is. I've tried calculating $g'$ to see if it is negative everywhere, but have failed to do so.
Suppose it is true. Then, since $f'(0)<0$, the function $x\mapsto e^{f'(0)x}$ is integrable on $x>0$, so we have found a dominating function for the $f_n$.
Furthermore, we need to show that there is an integrable function $h$ to which the $f_n$ converge. I cannot see how to show this pointwise convergence.
Edit: I have found conclusive proofs for my two problems and have added them as an answer.
To conclude the problem, the limit is $$\lim_{n\to\infty}\int_0^nf\left(\frac{x}{n}\right)^ndx = \int_0^\infty e^{-|f'(0)|x}dx = \dfrac{1}{|f'(0)|}.$$
Best Answer
Proof:
For $x=0$, this is true since $f(0)=1$. Let $x>0$. We whish to show that $$\left|f\left(\dfrac{x}{n}\right)^n - e^{f'(0)x} \right|\to 0.$$ This is equivalent to $$\dfrac{f\left(\dfrac{x}{n}\right)^n }{e^{f'(0)x}}\to 1.$$ (Note that both denominator and enumerator are positive) Which on its turn is equivalent to $$d_n =\ln \dfrac{f\left(\dfrac{x}{n}\right)^n }{e^{f'(0)x}} \to 0.$$ We will show that this is indeed true. Note that $$d_n = n\ln f\left(\dfrac{x}{n}\right)-f'(0)x$$ and hence that (we took $x>0$) $$\dfrac{d_n}{x} = g\left(\dfrac{x}{n}\right)-f'(0).$$
We already know that $g(0) = f'(0)$, so then $d_n \to 0$.
Proof:
Note that from what we know about $f$, it must be true that for all $y\in [0,1]: f(y)\leq f(0)=1$. Then it is also true that for all $y\in[0,1]:$ $$\ln f(y) \leq \ln f(0).$$ Now, it is also true that for all $y\in (0,1]:$ $$\dfrac{\ln f(y)}{y} \leq \dfrac{\ln f(0)}{y}.$$ Now, note that $\ln f(0) = 0$, so for any $\varepsilon > 0$ we get that $$\dfrac{\ln f(y)}{y} \leq \dfrac{\ln f(0)}{y-\varepsilon}.$$ Taking the limit $\varepsilon \to y$ makes the right hand side $g(0) = f'(0)$.