absolute convergence of integral and boundedness of derivative

Theorem.  Assume that we have an http://planetmath.org/node/11865absolutely converging integral

$${\int }_a^{\mathrm{\infty }}f(x)𝑑x$$

where the real function $f$ and its derivative $f^{\prime }$ are continuous and $f^{\prime }$ additionally bounded on the interval $[a,\mathrm{\infty })$.  Then

$\underset{x\to \mathrm{\infty }}{lim}f(x)=\mathrm{\hspace{0.33em}0}.$(1)

Proof.  If  $c>a$,  we obtain

$${\int }_a^{c}f(x)f^{\prime }(x)dx=\frac{1}{2}\underset{a}{\overset{c}{/}}{(f(x))}^2=\frac{{(f(c))}^2-{(f(a))}^2}{2},$$

from which

${(f(c))}^2={(f(a))}^2+2{\displaystyle {\int }_a^c}f(x)f^{\prime }(x)𝑑x.$(2)

Using the boundedness of $f^{\prime }$ and the absolute convergence, we can estimate upwards the integral

$${\int }_a^c|f(x)f^{\prime }(x)|𝑑x={\int }_a^c|f(x)||f^{\prime }(x)|𝑑x\leqq M{\int }_a^c|f(x)|𝑑x\leqq M{\int }_a^{\mathrm{\infty }}|f(x)|𝑑x\mathit{\hspace{1em}}\forall c\in [a,\mathrm{\infty })$$

whence ${\int }_a^{\mathrm{\infty }}|f(x)f^{\prime }(x)|𝑑x$ is finite and thus ${\int }_a^{\mathrm{\infty }}f(x)f^{\prime }(x)𝑑x$ converges absolutely.  Hence (2) implies

$$\underset{c\to \mathrm{\infty }}{lim}{(f(c))}^2={(f(a))}^2+2{\int }_a^{\mathrm{\infty }}f(x)f^{\prime }(x)𝑑x,$$

i.e. $\underset{x\to \mathrm{\infty }}{lim}{(f(x))}^2$ exists as finite, therefore also

$$\underset{x\to \mathrm{\infty }}{lim}|f(x)|:=A.$$

Antithesis:  $A>0$.  It implies that there is an $x_0(\geqq a)$ such that

$$|f(x)|\geqq \frac{A}{2}\mathit{\hspace{1em}}\forall x\geqq x_0.$$

If now  $b>x_0$, then we had

$${\int }_{x_0}^b|f(x)|𝑑x\geqq \frac{A}{2}(b-x_0)⟶\mathrm{\infty }\mathit{\hspace{1em}}\text{as}b\to \mathrm{\infty }.$$

This means that ${\int }_{x_0}^{\mathrm{\infty }}|f(x)|𝑑x$ and consequently also ${\int }_a^{\mathrm{\infty }}|f(x)|𝑑x$ would be divergent.  Since it is not true, we infer that  $A=0$,  i.e. that the assertion (1) is true.