I have this problem in real analysis. I think it needs integral factor or knowledge of ODE to prove, but not sure how to it. Here is the question:

Let $f$ be a real valued continuous function on $[0,\infty]$ such that $$ \lim \limits_{x\to\infty}\left(f(x)+\int_{0}^{x}f(t)dt\right) $$ exists. Prove that $$ \lim \limits_{x\to\infty}f(x)=0 $$


Let $g(x)=e^x\int_{0}^{x}f(t)dt$, then $$ g'(x)=e^x\left(f(x)+\int_{0}^{x}f(t)dt\right) $$ So there is $$ \lim \limits_{x\to\infty}g'(x)e^{-x}=\lim \limits_{x\to\infty}\left(f(x)+\int_{0}^{x}f(t)dt\right)=A $$ So for any $\epsilon>0$, there is $M>0$, such that for all $x>M$ $$ A-\epsilon<g'(x)e^{-x}<A+\epsilon \hspace{5 mm} \text{or} \hspace{5 mm} (A-\epsilon)e^{x}<g'(x)<(A+\epsilon)e^{x} $$ By integrating on both side from $M$ to $x$, there is $$ (A-\epsilon)\left(e^{x}-e^M\right)<g(x)-g(M)<(A+\epsilon)\left(e^{x}-e^M\right) $$ So we have $$ \left((A-\epsilon)\left(e^{x}-e^M\right)+g(M)\right)e^{-x}<\int_{0}^{x}f(t)dt<\left((A+\epsilon)\left(e^{x}-e^M\right)+g(M)\right)e^{-x} $$ And $$ \varlimsup\limits_{x\to\infty}\int_{0}^{x}f(t)dt\leqslant \varlimsup\limits_{x\to\infty}\left((A+\epsilon)\left(e^{x}-e^M\right)+g(M)\right)e^{-x}=A+\epsilon $$ $$ \varliminf\limits_{x\to\infty}\int_{0}^{x}f(t)dt\geqslant \varliminf\limits_{x\to\infty}\left((A-\epsilon)\left(e^{x}-e^M\right)+g(M)\right)e^{-x}=A-\epsilon $$ So we have $$ 0\leqslant \varlimsup\limits_{x\to\infty}\int_{0}^{x}f(t)dt-\varliminf\limits_{x\to\infty}\int_{0}^{x}f(t)dt\leqslant 2\epsilon $$ Since ϵ is arbitrary, we have $$ \varlimsup\limits_{x\to\infty}\int_{0}^{x}f(t)dt=\varliminf\limits_{x\to\infty}\int_{0}^{x}f(t)dt=A \hspace{5 mm} $$ Or $$ \lim\limits_{x\to\infty}\int_{0}^{x}f(t)dt=\lim \limits_{x\to\infty}\left(f(x)+\int_{0}^{x}f(t)dt\right) $$ So finally we have $$ \lim \limits_{x\to\infty}f(x)=\lim \limits_{x\to\infty}\left(\left(f(x)+\int_{0}^{x}f(t)dt\right)-\int_{0}^{x}f(t)dt\right)=0 $$


We are given $f(x) + \int_0^x f \to L.$ To show $f(x)\to 0,$ we need to show $ \int_0^x f \to L.$ But note

$$\tag 1 \int_0^x f =\frac{e^x\int_0^x f}{e^x}.$$

Since the denominator on the right $\to \infty,$ we can contemplate using L'Hopital. Let's try it: The quotient of derivatives is

$$\tag 2 \frac{e^x(f(x)+\int_0^x f)}{e^x} = f(x)+\int_0^x f.$$

The right hand side of $(2)\to L$ by hypothesis. Therefore, by L'Hopital, $(1)\to L,$ and we are done.

  • $\begingroup$ It's funny: I gave an answer to this four years ago. Today I found it a little confusing! So I edited it to the above. $\endgroup$ – zhw. Sep 17 at 19:04

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