Let \(f\) be a continuous real-valued function on \([0,1]\) such that, for each \(x_0\in[0,1)\), \[ \limsup_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}\ge 0. \] Prove that \(f\) is nondecreasing.
Solution
1. Suppose, for contradiction, that there exist \(0\leq a<b\leq1\) such that\[ f(a)>f(b). \]
This is precisely the negation of the assertion that \(f\) is nondecreasing.
2. There is an \(\varepsilon>0\) such that the continuous function\[ g(x)=f(x)+\varepsilon x \]satisfies \(g(a)>g(b)\) and\[ \limsup_{x\to x_0^+}\frac{g(x)-g(x_0)}{x-x_0}\geq\varepsilon>0 \]for every \(x_0\in[0,1)\).
By step 1, \[ \frac{f(a)-f(b)}{b-a}>0. \] Choose \[ 0<\varepsilon<\frac{f(a)-f(b)}{b-a}. \] Then \[ g(a)-g(b) =f(a)-f(b)-\varepsilon(b-a)>0. \] For every \(x_0\in[0,1)\), \[ \frac{g(x)-g(x_0)}{x-x_0} = \frac{f(x)-f(x_0)}{x-x_0}+\varepsilon. \] Taking the upper limit as \(x\to x_0^+\) and using the hypothesis on \(f\) gives \[ \limsup_{x\to x_0^+}\frac{g(x)-g(x_0)}{x-x_0} = \limsup_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}+\varepsilon \geq\varepsilon>0. \]
3. The function \(g\) has a point \(c\in[a,b)\) such that\[ \limsup_{x\to c^+}\frac{g(x)-g(c)}{x-c}\leq0. \]
Since \(g\) is continuous, it attains a maximum on the compact interval \([a,b]\). Let \(c\) be a point where this maximum is attained. Step 2 gives \(g(a)>g(b)\), so \(c\neq b\) and hence \(c\in[a,b)\).
For every \(x\in(c,b]\), \[ g(x)\leq g(c). \] Thus \[ \frac{g(x)-g(c)}{x-c}\leq0 \] for all \(x>c\) sufficiently close to \(c\). Taking the upper limit proves the claim.
4. The assumption in step 1 is impossible.
Since \(c<b\leq1\), one has \(c\in[0,1)\). Step 2 applied at \(x_0=c\) gives \[ \limsup_{x\to c^+}\frac{g(x)-g(c)}{x-c}\geq\varepsilon>0, \] contradicting step 3.
5. The function \(f\) is nondecreasing on \([0,1]\).
By step 4, there do not exist \(a<b\) with \(f(a)>f(b)\). Hence for every \(a<b\) in \([0,1]\) one has \(f(a)\leq f(b)\).
6. Q.E.D.
Step 5 is the required conclusion.