Let \(f\) be a real-valued continuous function on \([0,\infty)\) such that \[ \lim_{x\to\infty}\left(f(x)+\int_0^x f(t)\,dt\right) \] exists. Prove that \[ \lim_{x\to\infty}f(x)=0. \]
Solution
Define \[ F(x)=\int_0^x f(t)\,dt \] and \[ g(x)=f(x)+F(x). \] Let \[ L=\lim_{x\to\infty}g(x). \]
1. The function \(F\) satisfies\[ F'(x)+F(x)=g(x), \qquad F(0)=0. \]
Since \(f\) is continuous, the fundamental theorem of calculus gives \[ F'(x)=f(x). \] Therefore the definition of \(g\) gives \[ F'(x)+F(x)=f(x)+F(x)=g(x), \] and the defining integral gives \(F(0)=0\).
2. For every \(x\geq0\),\[ F(x)=e^{-x}\int_0^x e^t g(t)\,dt. \]
Multiplying the equation in step 1 by \(e^x\) gives \[ \bigl(e^xF(x)\bigr)'=e^xg(x). \] Integrating from \(0\) to \(x\) and using \(F(0)=0\) yields \[ e^xF(x)=\int_0^x e^t g(t)\,dt, \] which is the asserted formula.
3. One has\[ \lim_{x\to\infty}F(x)=L. \]
By step 2, \[ \begin{aligned} F(x)-L &= e^{-x}\int_0^x e^t\bigl(g(t)-L\bigr)\,dt -Le^{-x}. \end{aligned} \] Let \(\varepsilon>0\). Since \(g(t)\to L\), choose \(T\) such that \[ \abs{g(t)-L}<\varepsilon \] for all \(t\geq T\). For \(x\geq T\), \[ \begin{aligned} \abs{F(x)-L} &\leq e^{-x}\int_0^T e^t\abs{g(t)-L}\,dt + e^{-x}\int_T^x e^t\varepsilon\,dt + \abs L e^{-x}\\ &\leq C_Te^{-x}+\varepsilon+\abs L e^{-x}, \end{aligned} \] where \[ C_T=\int_0^T e^t\abs{g(t)-L}\,dt<\infty. \] Taking the upper limit as \(x\to\infty\) gives \[ \limsup_{x\to\infty}\abs{F(x)-L}\leq\varepsilon. \] Since \(\varepsilon\) is arbitrary, \(F(x)\to L\).
4. One has\[ \boxed{\lim_{x\to\infty}f(x)=0}. \]
By definition, \[ f(x)=g(x)-F(x). \] The hypothesis gives \(g(x)\to L\), and step 3 gives \(F(x)\to L\). Therefore \[ f(x)\longrightarrow L-L=0. \]
5. Q.E.D.
Step 4 is the required conclusion.