Let \(f_n:[0,1]\to[0,\infty)\) be continuous and suppose \[ f_1(x)\ge f_2(x)\ge f_3(x)\ge\cdots \] for every \(x\in[0,1]\). Set \[ f(x)=\lim_{n\to\infty}f_n(x), \qquad M=\sup_{0\le x\le1}f(x). \]
- Prove that there is \(t\in[0,1]\) such that \(f(t)=M\).
- Give an example showing that part 1 can fail if the monotonicity is only eventual pointwise: for each \(x\) there is \(n_x\) such that \(f_n(x)\ge f_{n+1}(x)\) for all \(n\ge n_x\).
Solution
1. There exists \(t\in[0,1]\) such that \(f(t)=M\).
For each positive integer \(k\), choose \(x_k\in[0,1]\) such that \[ f(x_k)>M-\frac1k. \] By compactness of \([0,1]\), some subsequence \((x_{k_j})\) converges to a point \(t\in[0,1]\).
Fix a positive integer \(n\). Since the sequence \((f_m(x))_m\) decreases to \(f(x)\) at every \(x\), \[ f_n(x_{k_j})\ge f(x_{k_j})>M-\frac1{k_j}. \] Letting \(j\to\infty\) and using continuity of \(f_n\) gives \[ f_n(t)\ge M. \] This holds for every \(n\). Therefore \[ f(t) =\lim_{n\to\infty}f_n(t) \ge M. \] By the definition of \(M\) as the supremum of \(f\), also \(f(t)\le M\). Hence \(f(t)=M\).
2. For part 2, define\[ f_n(x)\coloneqq x\min\{1,n(1-x)\}, \qquad 0\le x\le1. \]Then each \(f_n\) is continuous and nonnegative.
Both functions \(1\) and \(n(1-x)\) are continuous, so their pointwise minimum is continuous. Multiplication by \(x\ge0\) preserves continuity and nonnegativity.
3. The sequence from step 2 is eventually nonincreasing atevery point.
Fix \(x<1\). For every integer \[ n\ge\frac1{1-x}, \] we have \(n(1-x)\ge1\), so \[ f_n(x)=x. \] Thus the sequence is eventually constant, hence eventually nonincreasing, at every \(x<1\). At \(x=1\), \[ f_n(1)=0 \] for every \(n\), so the same is true there.
4. The limiting function in this example does not attain itssupremum.
By step 3, \[ f(x)\coloneqq\lim_{n\to\infty}f_n(x) = \begin{cases} x,&0\le x<1,\\ 0,&x=1. \end{cases} \] Thus \[ \sup_{0\le x\le1}f(x)=1, \] but \(f(x)<1\) for every \(x\in[0,1]\). Hence the supremum is not attained.
5. Q.E.D.
Step 1 proves part 1, and steps <1>2–<1>4 provide the required counterexample for part 2.