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Chapter 8: Problem 20

Give an example of a decreasing sequence \(\left(f_{n}\right)\) of continuous functions on \([0,1)\) that converges to a continuous limit function, but the convergence is not uniform on \([0,1)\).

Short Answer

Expert verified
The sequence of functions \(f_{n}(x) = nx(1-x)^n\) on \([0,1)\) converges to the limit function \(f(x) = 0\) for all \(x\) in \([0,1)\), but the convergence is not uniform because the maximum value of each \(f_{n}\) is not going to zero as \(n\) goes to infinity.

Step by step solution

01

Define the sequence of functions

An example of a sequence of such functions is \(f_{n}(x) = nx(1-x)^n\) for \(x\) in \([0,1)\). Each \(f_{n}(x)\) is clearly continuous on \([0,1)\).
02

Determine the limit function

To find the limit function, take the limit of \(f_{n}(x)\) as \(n\) approaches infinity. The limit function is then \(f(x) = \lim_{n \to \infty} {f_{n}(x)}\). By applying L'Hopital's rule, we find that the limit function is \(f(x) = 0\) for all \(x\) in \([0,1)\).
03

Prove non-uniform convergence

To show that the convergence is not uniform on \([0,1)\), find the maximum of each \(f_{n}\) by setting its derivative to zero and solving for \(x\). The derivative of \(f_{n}(x)\) is \(f_{n}^{'}(x) = n(1-x)^{n-1} - n^{2}x(1-x)^{n-1}\). Setting it to zero and solving for \(x\) gives \(x = 1/(n+1)\). Substituting this back into \(f_{n}(x)\) gives \(f_{n}(x) = 1/e\) for all \(n\). Therefore, the maximum value of each \(f_{n}\) is not going to zero as \(n\) goes to infinity, which is a requirement for uniform convergence.

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Most popular questions from this chapter

Suppose \(\left(f_{n}\right)\) is a sequence of continuous functions on an interval \(I\) that converges uniformly on \(I\) to a function \(f .\) If \(\left(x_{n}\right) \subseteq I\) converges to \(x_{0} \in I\), show that \(\lim \left(f_{n}\left(x_{n}\right)\right)=f\left(x_{0}\right)\).

Show that \(\lim \left(x^{2} e^{-n x}\right)=0\) and that \(\lim \left(n^{2} x^{2} e^{-n x}\right)=0\) for \(x \in \mathbb{R}, x \geq 0\)

Let \(a_{k}>0\) for \(k=1, \cdots, n\) and let \(A:=\left(a_{1}+\cdots+a_{n}\right) / n\) be the arithmetic mean of these numbers. For each \(k\), put \(x_{k}:=a_{k} / A-1\) in the inequality \(1+x \leq e^{x}(\) valid for \(x \geq 0)\). Multiply the resulting terms to prove the Arithmetic-Geometric Mean Inequality (6) $$ \left(a_{1} \cdots a_{n}\right)^{1 / n} \leq \frac{1}{n}\left(a_{1}+\cdots+a_{n}\right) $$ Moreover, show that equality holds in \((6)\) if and only if \(a_{1}=a_{2}=\cdots=a_{n}\).

Show that the sequence \(\left(x^{2} e^{-n x}\right)\) converges uniformly on \([0, \infty)\).

Let \(f_{n}(x):=n x /\left(1+n x^{2}\right)\) for \(x \in A:=[0, \infty)\), Show that each \(f_{n}\) is bounded on \(A_{1}\) but the pointwise limit \(f\) of the sequence is not bounded on \(A\). Does \(\left(f_{n}\right)\) converge uniformly to \(f\) on \(A\) ?
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