Prove the following theorem, or find a counterexample: If \(p\) and \(q\) are continuous real-valued functions on \(\mathbb R\) such that \(|q(x)|\le|p(x)|\) for all \(x\), and if every solution \(f\) of the differential equation \[ f'+qf=0 \] satisfies \(\lim_{x\to+\infty}f(x)=0\), then every solution \(f\) of the differential equation \[ f'+pf=0 \] satisfies \(\lim_{x\to+\infty}f(x)=0\).
Solution
1. Take\[ \boxed{q(x)=1,\qquad p(x)=-1} \]for every \(x\in\RR\).
Both functions are continuous, and for every \(x\in\RR\), \[ \abs{q(x)}=1=\abs{p(x)}. \] Thus the required pointwise inequality \(\abs{q(x)}\leq\abs{p(x)}\) holds.
2. Every solution of\[ f'+qf=0 \]tends to \(0\) as \(x\to+\infty\).
For the choice in step 1, the equation is \[ f'+f=0. \] Multiplying by \(e^x\) gives \[ \bigl(e^x f(x)\bigr)'=0, \] so every solution has the form \[ f(x)=Ce^{-x} \] for some constant \(C\). Hence \(\lim_{x\to+\infty}f(x)=0\).
3. The corresponding equation\[ f'+pf=0 \]has a solution that does not tend to \(0\) as \(x\to+\infty\).
For the choice in step 1, this equation is \[ f'-f=0. \] The function \[ f(x)=e^x \] is a solution, and \(\lim_{x\to+\infty}e^x=+\infty\). Thus this solution does not tend to \(0\).
4. Q.E.D.
Steps 1 and 2 verify all hypotheses of the proposed theorem, while step 3 shows that its conclusion fails. Therefore the boxed pair in step 1 is a counterexample.