# Calculate $\lim_{n\to\infty}\int_0^nf(\frac{x}{n})^ndx$.

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 $$x\in(0,1]:0. Calculate $$\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 $$\lim_{n\to\infty}\int_0^nf\left(\frac{x}{n}\right)^ndx = \int_0^\infty e^{-|f'(0)|x}dx = \dfrac{1}{|f'(0)|}.$$

• @LightYagami I agree that the answer is 1. However, plugging in a function $f$ is hardly a proof ;). I've found a proof and am writing an answer as we speak. Jan 7, 2021 at 14:08
• That's nice. I will be happy to read it.
– V.G
Jan 7, 2021 at 14:10
• @LightYagami sorry, the limit is actually $|1/f'(0)|$. Since the limiting function is $e^{f'(0)x}$. Jan 7, 2021 at 14:15

For all $$x\ge 0$$: $$f\left(\dfrac{x}{n}\right)^n\to e^{f'(0)x}.$$

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 $$d_n \to 0$$.

For all $$y\in[0,1]:$$ $$g(y) \leq f'(0).$$

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 $$g(0) = f'(0)$$.

• Thats correct !
– EDX
Jan 7, 2021 at 14:14

Here is an alternative proof that doesn't require convergence theorems for integrals. Write $$\ell=|f'(0)|$$ and take $$0<\epsilon<\frac12\ell$$: the factor $$1/2$$ will come into play at the end and is purely here for convenience. Since $$f$$ is differentiable at $$0$$ there exists $$\eta\in(0,1]$$ such that $$\forall x\in [0,\eta], \quad |f(x)-1+\ell x|\leq\epsilon x$$ Without loss of generality we further assume $$\eta < \frac{1}{2\ell}$$. Then for all $$x\in[0,\eta]$$, $$0<1-\ell x-\epsilon x\leq f(x)\leq 1-\ell x+\epsilon x$$ and so $$\begin{array}{rcl} \displaystyle \int_0^nf(x/n)^n~dx &=& \displaystyle n\int_0^1f(x)^n~dx\\ &=& \displaystyle \underbrace{n\int_0^\eta f(x)^n~dx}_{I_n} + \underbrace{n\int_\eta^1f(x)^n~dx}_{II_n} \end{array}$$ We show that the first term $$I_n$$ converges to $$\frac1{f'(0)}$$ and the second term $$II_n$$ tends to $$0$$.

The second term. This is swiftly dealt with: by continuity of $$f$$ and $$0 on $$(0,1]$$ it follows that $$M=\max_{[\eta,1]}f<1$$. Then $$0\leq II_n\leq nM^n$$ so that $$II$$ indeed (converges and) tends to $$0$$.

The first term. Integrating the inequality from above yields $$-n\bigg[\frac{(1-(\ell+\epsilon)x)^{n+1}}{(n+1)(\ell + \epsilon)}\bigg]^\eta_0 \leq I_n\leq -n\bigg[\frac{(1-(\ell - \epsilon)x)^{n+1}}{(n+1)(\ell - \epsilon)}\bigg]^\eta_0$$ The LHS equals $$\frac{n}{n+1}\frac{1}{\ell+\epsilon} - C_nq_1^n=\frac{1}{\ell+\epsilon}+o(1)$$ for $$C_n=\frac{n}{(n+1)(\ell + \epsilon)}$$ and $$q_1=1-(\ell + \epsilon)\eta\in(0,1)$$.

Similarly, the RHS equals $$\frac{n}{n+1}\frac{1}{\ell-\epsilon} - D_nq_2^n=\frac{1}{\ell-\epsilon}+o(1)$$ for $$D_n=\frac{n}{(n+1)(\ell - \epsilon)}$$ and $$q_2=1-(\ell - \epsilon)\eta\in(0,1)$$.

Since $$\frac{1}{\ell+2\epsilon}<\frac{1}{\ell+\epsilon}$$ and $$\frac{1}{\ell-\epsilon}<\frac{1}{\ell-2\epsilon}$$, there exists an integer $$n_0$$ such that for all $$n\geq n_0$$ $$\frac{1}{\ell+2\epsilon} \leq I_n \leq \frac{1}{\ell-2\epsilon}$$ This establishes the convergence of $$I_n$$ to $$\frac1\ell=\frac1{|f'(0)|}$$ and finishes the proof.

• That's a lot of estimations and inequalities. Thanks for the proof! Jan 7, 2021 at 15:43