Let $f\in\mathcal{C}^1(\mathbb{R}_+,\mathbb{R})$ s.t. $\displaystyle \int_0^{+\infty} f(t)\cdot dt$ and $\displaystyle \int_0^{+\infty} f'^2(t)\cdot dt$ converge. Show $f$ is bounded.

My question is : if I use only the following statements :
$f\in\mathcal{C}^1(\mathbb{R}_+,\mathbb{R})$ and $\displaystyle \int_0^{+\infty} f(t)\cdot dt$ converges, it is enough to conclude $f$ is bounded, indeed $f(0)$ exists, and $f(t)\underset{t\to +\infty}\to 0$ then with the extreme value theorem and the definition of the limit we can show $f$ is bounded.

Now if modify the statement of this problem

Let $f\in\mathcal{C}^1(\mathbb{R}_+^*,\mathbb{R})$ s.t. $\displaystyle \int_0^{+\infty} f(t)\cdot dt$ and $\displaystyle \int_0^{+\infty} f'^2(t)\cdot dt$ converge. Show $f$ is bounded.

Let $A:=\displaystyle \int_0^{+\infty} f'^2(t)\cdot dt$ and $\varepsilon>0$

$\displaystyle \int_x^y f'(t)\cdot dt =f(x)-f(y)$


$|f(x)-f(y)|=\left|\displaystyle \int_x^y f'(t)\cdot dt\right|\le\left|\displaystyle \int_x^y f'^2(t)\cdot dt\right|^{1/2}\cdot\quad\left|\displaystyle \int_x^y \;\; dt\right|^{1/2}<\sqrt{A}\sqrt{y-x}$

So I choose $\eta = \dfrac{\varepsilon^2}{A}$ we have $\forall (x,y)\in \mathbb{R}^2, |x-y|<\eta \implies |f(x)-f(y)|<\varepsilon$ so $f$ is uniformly continuous on $\mathbb{R}_+^*$

Let $a<\eta$ for all $x\in(0,a) \quad f(a)-\varepsilon<f(x)<f(a)+\varepsilon$

With this same $\varepsilon$ there exists $c$ such that $x>c$ we have $-\varepsilon<f(x)<\varepsilon$ and with the extrem value theorem we have $m<f(x)<M$ on $[a,c]$

With $M'=\max(f(a)+\varepsilon,\varepsilon,M)$ and $m'=\min(f(a)-\varepsilon,-\varepsilon,m)$

We have for all $x\in(0,+\infty) \quad m'<f(x)<M'$ the proof is complete $\square$

Is that all right?

  • 5
    $\begingroup$ your first argument is false. even if $f$ is integrable and in $C^1$m it doesn't mean that it is bounded. In particular, it is false that $f$ goes to zero at infinity $\endgroup$ – Exodd Dec 16 '17 at 0:22
  • $\begingroup$ For a counter-example, take a function which is 0, but at every integer, it has an increasingly thin spike up to 1. If the width of the spikes decays sufficiently rapidly, this will still be integrable. $\endgroup$ – Alfred Yerger Dec 16 '17 at 0:23
  • $\begingroup$ @Exodd Thanks for example : $$f : x\to \sin(x^2)$$ $\endgroup$ – Stu Dec 16 '17 at 7:51
  • $\begingroup$ @Stu : Is it not supposed that $\int_0^{+\infty}|f(t)|dt $ converges ? It will be true with. $\endgroup$ – user371663 Dec 17 '17 at 15:13
  • $\begingroup$ @Lucas I don't think that will be true, see the answer of Alfred Yerger $\endgroup$ – Stu Dec 17 '17 at 22:56

It is sufficient to show $f$ tends to $0$ to prove it is bounded due to continuity on $[0,+\infty($ $(1)$. And for that since $f$ is integrable, it is sufficient to show it is uniformly continuous on $[0,+\infty($ $(2)$. It is a consequence of the inequality with $|f(x)-f(y)|$ in your post.

$(1)$ is pretty obvious. I let you $(2)$ to justify. Note that all kind of triangle functions are no longer valid as counter-example since you have proved $f$ is uniformly continuous.

Here is a solution :

$(1)$ is a easy consequence of preserving compactness by continuity. For $(2)$, suppose $f$ does not tend to $0$ in $+\infty$ i.e. $\exists \epsilon, \forall A \in [0,+\infty(, \ \exists \ x≥A, |f(x)|≥ \epsilon$ $(*)$

But $f$ is uniformly continous on $[0,+\infty($, so $\exists \eta, \ \forall x,y \in [0,+\infty(, \ |x-y|≤\eta \implies |f(x)-f(y)|≤\epsilon/2$ $(**)$

Now, use $(*)$ to create a sequence $(x_n)$ that tends to $+\infty$ satisfying $\forall n, |f(x_n)|≥ \epsilon$

Then $(**)$ will contradict the fact $\int_{x_n}^{x_n + \eta}f(t)dt \to 0$ which is a consequence of $\int_{0}^{+\infty}f(t)dt$ converges.

Indeed, $\forall n$, on $[x_n, x_n + \eta], |f|≥\epsilon/2$ so by continuity, $f≥\epsilon/2$ or $f≤-\epsilon/2$ hence $\forall n, |\int_{x_n}^{x_n + \eta}f(t)dt| ≥\eta *\epsilon/2 > 0$

  • $\begingroup$ Actually, it was not necessary that $\int_0^{+\infty}|f(t)|dt$ converges. $\endgroup$ – user371663 Dec 18 '17 at 4:07
  • $\begingroup$ Note that all the problem was to prove $f \to 0$, it was not obvious a priori since it is very false in general. $\endgroup$ – user371663 Dec 18 '17 at 5:46

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