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I'm looking for a good proof for the following problem. I think it's pretty intuitive why this is true, but I'm having trouble expressing it rigorously.

Suppose that $f$ is twice differentiable and that $f''$ is bounded on $(0, \infty)$. Furthermore, suppose that $\lim_{x\to \infty} f(x)=0$. Prove that $\lim_{x\to\infty} f'(x)=0$.

I've already tried writing out the Taylor expansion for $f$ and tweaking that, but I haven't had much success. Any insight would be greatly appreciated.

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  • $\begingroup$ Have you done anything yet? $\endgroup$ – 5xum Apr 28 '14 at 7:10
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    $\begingroup$ Question has been updates. Sorry, I saved too soon! $\endgroup$ – Shangchen Yong Apr 28 '14 at 7:42
  • $\begingroup$ It's not very intuitive IMO. For example if you only ask for piecewise differentiable, then the result is false... $\endgroup$ – Najib Idrissi Apr 28 '14 at 8:22
  • $\begingroup$ Reading the question only in the overview, the obvious reaction is " but that is wrong, take sin (x^2) / x ". Now you take the fact that f'' is bounded to exclude that kind of function. f'' bounded means it takes some time for f' to become > epsilon, and in that time f itself has grown enough to not converge -> 0 anymore. $\endgroup$ – gnasher729 Apr 28 '14 at 10:20
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I think you can get there with Taylor.

Firstly, since $\lim_{x\to\infty} f(x)=0$ Then for any $\epsilon > 0$ you can find X such that $|f(x)| < \epsilon$ for all $x \ge X$, and furthemore, you can find X such that for all X' > X you have $|f(X) - f(X')| < \epsilon$. And, for any given $\kappa$ with $X'-X \ge \kappa$ you can find X such that for all $X' \ge X + \kappa$ you have $|f(X) - f(X')| < \epsilon\kappa < \epsilon(X'-X)$

For such X and X', take the taylor expansion of $f(X')$ from X, i.e. $f(X') = f(X) + (X'-X)f'(X) + (X'-X)^2f''(z)/2$, where $X \le z \le X'$, then $|f(X') - f(X)| = |(X'-X)f'(X) + (X'-X)^2f''(z)/2| < \epsilon$(X'-X), and using the triangle rule for modulus inequality,

$|(X'-X)f'(X)| < |(X'-X)^2f''(z)/2| + \epsilon$(X'-X).

For $X' - X > \kappa$ we can divide through so that

$|f'(X)| < (X'-X)|f''(z)/2| + \epsilon$, and as $f''$ is bounded, say $|f′′(z)| < 2M$,

$|f′(X)|<(X′−X)M+\epsilon$. Now let X' = X + $\kappa$ so that

$|f′(X)|<\kappa M+\epsilon$

Since we can make $\epsilon$ and $\kappa$ as small as we like, then $\lim_{x\to\infty} f'(x)=0$

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Apply Landau's inequality on the tail of $(0,+\infty)$, i.e. $(M,+\infty)$ with $M \gg 1$: $$ |f'(x)| \leq \|f''\|_{L^\infty(M,+\infty)} \|f\|_{L^\infty(M,+\infty)} \leq \varepsilon \|f''\|_{L^\infty(M,+\infty)} $$ because $|f(x)|<\varepsilon$ whenever $x>M$.

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  • $\begingroup$ Unfortunately, we aren't allowed to use that as we haven't covered it. This is for a first course in real analysis. $\endgroup$ – Shangchen Yong Apr 28 '14 at 8:42
  • $\begingroup$ You can also find the complete solution to your problem in E. Landau's original paper (1913): dx.doi.org/10.1112/plms/s2-13.1.43 $\endgroup$ – Siminore Apr 29 '14 at 11:04
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What I'm about to say may well be included in the answers above, but I found them a bit hard to read. A classic result that follows from Taylor's Theorem is this: If $|f|\le M_0$ and $|f''|\le M_2$ on $(0,\infty)$, then for any $x$, we have $$|f'(x)|\le \frac 2h M_0 + \frac h2 M_2 \qquad\text{for any } h>0.$$ By minimizing over all $h>0$, you get a nice upper bound on $|f'|$.

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Let $M$ be an upper bound of $|f''(x)|$ on $(0,\infty)$. For any $\varepsilon > 0$, let $\displaystyle \delta = \frac{\varepsilon}{3M}$. Since $\lim\limits_{x\to\infty} f(x) = 0$, there exists a $X$ such that $\displaystyle |f(x)| < \frac{\varepsilon\delta}{3}$ whenever $x > X$. For such a $x$, apply MVT to $f(x)$ on $(x,x+\delta)$, we can find a $\eta_1 \in (0,1)$ such that $$f'(x + \eta_1 \delta) = \frac{f(x+\delta)-f(x)}{\delta}$$ Apply MVT to $f'(x)$ on $(x,x+\eta_1\delta)$, we can find another $\eta_2 \in (0,1)$ such that $$f''(x+ \eta_1\eta_2\delta) = \frac{f'(x+\eta_1\delta)-f'(x)}{\eta_1\delta}$$ In terms of $\eta_1, \eta_2$, we can rewrite $f'(x)$ as

$$f'(x) - f'(x+\eta_1 \delta) + \frac{f(x+\delta)-f(x)}{\delta} = - f''(x+\eta_1\eta_2\delta) \eta_1\delta + \frac{f(x+\delta)-f(x)}{\delta}$$ As a result, we can bound $|f'(x)|$ over $(X,\infty)$ as $$ |f'(x)| \le |f''(x+\eta_1\eta_2\delta)|\eta_1\delta + \frac{|f(x+\delta)| + |f(x)|}{\delta} \le M\delta + \frac{1}{\delta}\left(\frac{\varepsilon\delta}{3} +\frac{\varepsilon\delta}{3}\right) = \varepsilon $$

Since $\varepsilon$ is arbitrary, this leads to $\lim\limits_{x\to\infty} f'(x) = 0$.

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For $\epsilon>0$, let $$I_1 \cup J_1 \cup K_1 \cup I_2 \cup J_2 \cup K_2 \cup \cdots \subset (0,+ \infty)$$ be a sequence of adjacent segments such that $f' \geq \epsilon$ on $I_s$, $|f'| < \epsilon$ on $J_s$ and $f' \leq - \epsilon$ on $K_s$.

If we suppose by contradiction that $f'(x)$ does not converge to zero as $x \to + \infty$, there exists $\epsilon>0$ such that the decomposition above satisfies $\mathrm{lg}(I_s) \neq 0$ or $\mathrm{lg}(K_s) \neq 0$ for infinitely many $s \geq 1$, and such that $|f'|$ attains $2\epsilon$ on $I_s$ and $K_s$ (if they are nonempty). Without loss of generality, we may suppose that $\mathrm{lg}(I_s) \neq 0$ for infinitely many $s \geq 1$ (otherwise, take $-f$).

On $I_s$, $|f| \leq \sup\limits_{Is} |f| =\eta_s$, $|f''|\leq K$ and $f' \geq \epsilon$. Now

$$2 \eta_s \geq f(b_s)-f(a_s)= \int_{a_s}^{b_s} f'(t)dt \geq \epsilon (b_s-a_s)$$ where $I_s=[a_s,b_s]$, hence $\mathrm{lg}(I_s)=b_s-a_s \leq \frac{2 \eta_s}{K}$ so $\mathrm{lg}(I_{s_k}) \underset{k \to + \infty}{\longrightarrow} 0$ for some subsequence.

Now if $x_{s_k} \in I_{s_k}$ is such that $f'(x_{s_k})=2 \epsilon$, then there exists $z_{s_k} \in (a_s,x_{s_k})$ such that $$|f''(z_{s_k})|= \frac{|f'(x_{s_k})-f'(a_s)|}{|x_{s_k}-a_s|}$$ hence $$\epsilon= |f'(x_{s_k})-f'(a_s)| \leq K |x_{s_k}-a_s| \leq K \mathrm{lg}(I_{s_k}) \underset{k \to + \infty}{\longrightarrow} 0,$$

a contradiction with $\epsilon>0$.

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I may have a proof using mainly MVT. The idea is that $f'$ is lipschitz, so it can't vary "too quick". If $f'$ doesn't go to zero at infinity, you will find by the Lipschitz property a neighboorhood (with a controllable size) around the points that are "far from zero", where $f'$ remains "big" enough. But then using the MVT you can show that implies that $f$ also increases "enough" and that will contradict the fact that, at infinity, the distance between two images of $f$ is going to zero. Here you are :

Since $f''$ is bounded, it means that, using MVT, $f'$ is Lipschitz (the constant is $sup|f''|$, let's call it $K$). Now, suppose that $f'$ is not going to $0$ at infinty. It means that there exists $\alpha > 0$, such that for any $A \geq 0$, there exists $x \geq A$ such that $|f'(x)| \geq \alpha$. Thus you can build a sequence (using $A=1$ to get an $x_1$, then $A=x_1+1$ to get $x_2$, and so on ...) $(x_n)$ that increases to $\infty$ such that $|f'(x_n)| \geq \alpha$

Now, $f'$ being lipschitz, you have that $$\forall x \in [x_n-\frac{\alpha}{4K},x_n + \frac{\alpha}{4K}], |f'(x)-f'(x_n)| \leq K|x-x_n| \leq K \frac{\alpha}{2K}=\frac{\alpha}{2}$$ which implies $$\forall x \in [x_n-\frac{\alpha}{4K},x_n + \frac{\alpha}{4K}], |f'(x)| \geq |f'(x_n)| - \frac{\alpha}{2} \geq \frac{\alpha}{2}$$ So, we just found that in a interval around each $x_n$, $f'$ is greater than a fixed value. But notice that neither this value ($\frac{\alpha}{2}$) nor the width of this interval ($\frac{\alpha}{2K}$) depends on $n$. Let's call this interval $I_n=[u_n,v_n]$.

For any $n \geq 0$, let's pick $u_n=x_n-\frac{\alpha}{4K},v_n=x_n+\frac{\alpha}{4K} \in I_n$. By the MVT, $|f(u_n) - f(v_n)| \geq sup_{I_n}|f'| \frac{\alpha}{2K} \geq \frac{\alpha^2}{4K}$. Once again, the bound doesn't depend on $n$...

Now, using the fact that $\lim f$ is $0$ at infinty, and $x_n$ goes to infinity (and so are $u_n$ and $v_n$), we can find a $n$ such that $|f(u_n)|$ and $|f(v_n)|$ are lesser than $\epsilon$ for a choosen $\epsilon > 0$. With $\epsilon=\frac{\alpha^2}{16K}$ (or anything strictly smaller than $\epsilon=\frac{\alpha^2}{8K}$) you will have : $$|f(u_n)-f(v_n)| \leq |f(u_n)| + |f(v_n)| \leq 2 \epsilon \leq \frac{\alpha^2}{8K} < \frac{\alpha^2}{4K}$$, which contradicts the previous point. Hence, $f'$ must go to zero at infinity.

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