Show $f$ is bounded 
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 $\varepsilon>0$
$\displaystyle \int_x^y f'(t)\cdot dt =f(x)-f(y)$ 
then 
$|f(x)-f(y)|=\left|\displaystyle \int_x^y f'(t)\cdot dt\right|\le\left|\displaystyle \int_x^y f'^2(t)\cdot dt\right|^{1/2}\cdot\quad\left|\displaystyle \int_x^y \;\; dt\right|^{1/2}<\sqrt{A}\sqrt{y-x}$
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?
 A: 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 Note that all kind of triangle functions are no longer valid as counter-example since you have proved $f$ is uniformly continuous.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 $\forall n, |\int_{x_n}^{x_n + \eta}f(t)dt| ≥\eta *\epsilon/2 > 0$
