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Main question:

Let $f : [0,infty) to mathbb R$ have an antiderivative. Furthermore, suppose $underset t to infty lim f(t) = 0$. Can I conclude that $f$ is bounded?

If $f$ were continuous, the answer would be “yes”, for obvious reasons. But, with just the given information, I find it difficult to tell.

Motivation:

This is given solely for context. Please do not answer the question here! I am trying to solve the following problem. Let $A$ be a square matrix whose eigenvalues all have negative real part. Let $f : mathbb R to mathbb C^n$ have an antiderivative, and suppose $underset t to infty lim f(t) = 0$. Show that

$$underset t to infty lim int_0^t e^(t-s)A g(s) ds = 0$$

NO. We can find a differentiable $F:Bbb Rto Bbb R$ with $F(x)=0$ for $xleq 0 $ and $F(x)=1$ for $xgeq 0,$ such that $F'(x): 0<x<1$ is unbounded. Then $f=F'$ has an anti-derivative, and $f'(x)=0$ for $xnot in (0,1),$ but $f$ is unbounded on $(0,1).$

(i). (Exercise).For $a<b$ and $c<d$ and for any $r>0$ there exists a monotonic differentiable $F:[a,b]to [c,d]$ with $F(a)=c$ and $F(b)=d$ and $F'(a)=F'(b)=0,$ such that $sup F'(x): xin (a,b)>r.$

(ii). For $nin Bbb N$ let $a_n=1-2^-n$ and $b_n=1-3^-n.$ By (i) there exists a monotonic differentiable $F:[a_n,a_n+1]to [b_n,b_n+1]$ with $F(a_n)=b_n$ and $F(a_n+1)=b_n+1 $ and $F'(a_n)=F'(a_n+1))=0$, such that $sup F'(x):xin (a_n,a_n+1)>n.$

And let $F(x)=0$ for $x<0$ and $F(y)=1$ for $ygeq 1.$

It remains to show that $F'(1)$ exists and is equal to $0.$ It suffices to show that $lim_xto 1^-frac 1-F(x)1-x=0.$ For $0<x<1$ there exists a unique $n_xin Bbb N$ such that $ a_n_xleq x<a_(n_x+1).$

So $0<frac 1-F(x)1-x<$ $ frac 1-F(a_n_x)1-a_(n_x+1)=$ $frac 3^-n_x2^-n_x-1,$ which $to 0$ as $n_xto infty.$ And $n_xto infty$ as $xto 1^-.$

Take

$$f(x) = begincases0, quad x = 0\ 2x sin frac1x^2 - frac2x cos frac1x^2, quad 0 < x leqslant1 endcases$$

and extend smoothly to $[0,infty)$ such that $f(x) to 0 $ as $x to infty$.

The function has an antiderivative with $F'(x) = f(x)$ such that on $[0,1]$ we have

$$F(x) = begincases0, quad x = 0 \ x^2 sinfrac1x^2, quad 0 < x leqslant 1 endcases$$

but $f$ is unbounded near $x = 0$.

No, one cannot so conclude. Let me add (a somewhat artificial) counterexample to the existing list:

Consider the hyperbolic branch $1/x$ in the first quadrant of $mathbb R^2$, and define the function $f$ with $f(x)=alpha$ for some $alphainmathbb R$ when $x=0$ and $f(x)=1/x$ for $xin(0,infty)$.

PS. You can only conclude that a function that satisfies your conditions is either bounded above or bounded below.