A function \(f:[0,1]\to\mathbb R\) is called upper semicontinuous if for every \(x\in[0,1]\) and every \(\varepsilon>0\) there exists \(\delta>0\) such that \[ |y-x|<\delta\quad\Longrightarrow\quad f(y)<f(x)+\varepsilon. \] Prove that an upper-semicontinuous function on \([0,1]\) is bounded above and attains its maximum at some point of \([0,1]\).
Solution
1. The function \(f\) is bounded above on \([0,1]\).
For each \(x\in[0,1]\), apply upper semicontinuity with \(\varepsilon=1\). There is an open neighborhood \(U_x\) of \(x\) in \([0,1]\) such that \[ y\in U_x \quad\Longrightarrow\quad f(y)<f(x)+1. \] The sets \(U_x\) cover the compact interval \([0,1]\), so finitely many of them, \[ U_{x_1},\ldots,U_{x_m}, \] already cover \([0,1]\). Set \[ B=\max_{1\le j\le m}\bigl(f(x_j)+1\bigr). \] For any \(y\in[0,1]\), choose \(j\) with \(y\in U_{x_j}\). Then \[ f(y)<f(x_j)+1\le B. \] Thus \(f\) is bounded above.
2. Let\[ M=\sup_{x\in[0,1]}f(x). \]There is a sequence \((x_n)\) in \([0,1]\) such that\[ f(x_n)>M-\frac1n \]for every \(n\ge1\).
Step 1 shows that \(M\) is finite. By the definition of supremum, for each \(n\) there is some \(x_n\in[0,1]\) with \(f(x_n)>M-1/n\).
3. Some subsequence \((x_{n_k})\) converges to a point\(p\in[0,1]\).
The interval \([0,1]\) is compact, so every sequence in it has a convergent subsequence whose limit remains in \([0,1]\).
4. One has\[ M\le f(p). \]
Fix \(\varepsilon>0\). By upper semicontinuity at \(p\), there is \(\delta>0\) such that \[ \abs{y-p}<\delta \quad\Longrightarrow\quad f(y)<f(p)+\varepsilon. \] Since \(x_{n_k}\to p\), for all sufficiently large \(k\), \[ f(x_{n_k})<f(p)+\varepsilon. \] On the other hand, step 2 gives \[ f(x_{n_k})>M-\frac1{n_k}. \] Letting \(k\to\infty\) yields \[ M\le f(p)+\varepsilon. \] Since \(\varepsilon>0\) was arbitrary, \(M\le f(p)\).
5. The function \(f\) attains its maximum at \(p\):\[ \boxed{f(p)=M}. \]
By definition of \(M\) as a supremum, \(f(p)\le M\). Step 4 gives the reverse inequality, so equality holds.
6. Q.E.D.
Step 1 proves boundedness above, and step 5 proves attainment of the maximum.