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

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  • $\begingroup$ @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. $\endgroup$ Jan 7 at 14:08
  • $\begingroup$ That's nice. I will be happy to read it. $\endgroup$ Jan 7 at 14:10
  • $\begingroup$ @LightYagami sorry, the limit is actually $|1/f'(0)|$. Since the limiting function is $e^{f'(0)x}$. $\endgroup$ Jan 7 at 14:15
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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 $g(0) = f'(0)$.

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  • $\begingroup$ Thats correct ! $\endgroup$
    – EDX
    Jan 7 at 14:14
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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 $0<f<1$ 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.

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    $\begingroup$ That's a lot of estimations and inequalities. Thanks for the proof! $\endgroup$ Jan 7 at 15:43

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