Let \(f:[0,1]\to\mathbb R\) be continuous with \[ f(0)=f(1)=0. \] Suppose \(f''\) exists on \((0,1)\) and \[ f''+2f'+f\ge0. \] Prove that \[ f(x)\le0 \] for every \(0\le x\le1\).
Solution
1. Define\[ g(x)\coloneqq e^x f(x). \]Then \(g\) is continuous on \([0,1]\), twice differentiable on \((0,1)\),\(g(0)=g(1)=0\), and\[ g''(x)\geq0 \]for \(0<x<1\).
The endpoint conditions give \[ g(0)=f(0)=0, \qquad g(1)=e f(1)=0. \] On \((0,1)\), \[ \begin{aligned} g''(x) &= e^x\bigl(f''(x)+2f'(x)+f(x)\bigr)\\ &\geq0, \end{aligned} \] because \(e^x>0\) and the assumed differential inequality holds.
2. The derivative \(g'\) is nondecreasing on \((0,1)\).
1. Hence \(g'(a)\leq g'(b)\).
3. For every \(x\in(0,1)\),\[ g(x)\leq0. \]
Fix \(x\in(0,1)\). The mean value theorem on \([0,x]\) gives \(c\in(0,x)\) such that \[ \frac{g(x)-g(0)}{x} = g'(c), \] and the mean value theorem on \([x,1]\) gives \(d\in(x,1)\) such that \[ \frac{g(1)-g(x)}{1-x} = g'(d). \] Since \(c<d\), step 2 gives \(g'(c)\leq g'(d)\). Using \(g(0)=g(1)=0\) from step 1, \[ \frac{g(x)}{x} \leq -\frac{g(x)}{1-x}. \] Multiplication by the positive number \(x(1-x)\) yields \[ (1-x)g(x)\leq-xg(x), \] and therefore \(g(x)\leq0\).
4. For every \(x\in[0,1]\),\[ \boxed{f(x)\leq0}. \]
For \(0<x<1\), step 3 and \[ f(x)=e^{-x}g(x) \] give \(f(x)\leq0\) because \(e^{-x}>0\). At the endpoints, \(f(0)=f(1)=0\) by hypothesis.
5. Q.E.D.
Step 4 is the required conclusion.