Let \(f,g:[0,1]\to[0,\infty)\) be continuous and suppose \[ \sup_{0\le x\le1}f(x)=\sup_{0\le x\le1}g(x). \] Prove that there exists \(t\in[0,1]\) such that \[ f(t)^2+3f(t)=g(t)^2+3g(t). \]
Solution
Set \[ M\coloneqq \max_{[0,1]}f = \max_{[0,1]}g. \] The maxima exist because \(f\) and \(g\) are continuous on the compact interval \([0,1]\).
1. There are points \(x_f,x_g\in[0,1]\) such that\[ (f-g)(x_f)\ge0, \qquad (f-g)(x_g)\le0. \]
Choose \[ f(x_f)=M, \qquad g(x_g)=M. \] Since \(g(x_f)\le M\), \[ (f-g)(x_f)=M-g(x_f)\ge0. \] Similarly, \(f(x_g)\le M\), so \[ (f-g)(x_g)=f(x_g)-M\le0. \]
2. There is \(t\in[0,1]\) such that\[ f(t)=g(t). \]
The function \(f-g\) is continuous. If either value in step 1 is zero, take the corresponding point. Otherwise the two values have opposite signs, so the intermediate value theorem gives a zero between \(x_f\) and \(x_g\).
3.\[ f(t)^2+3f(t)=g(t)^2+3g(t). \]
Substitute the equality \(f(t)=g(t)\) from step 2 into both sides.
4. Q.E.D.
Step 3 is the required conclusion.