(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For [tex]k = 1,2,\ldots[/tex] define [tex]f_k : [0,1] \to \mathbb{R}[/tex] by

[tex]

\begin{align*}

f_k(x) = \left\{

\begin{array}{ll}

4k^2x & 0 \leq \displaystyle x \leq \frac{1}{2k} \\

4k(1 - kx) & \displaystyle \frac{1}{2k} < x \leq \frac{1}{k} \\

0 & \displaystyle \frac{1}{k} < x \leq 1

\end{array}

\right.

\end{align*}

[/tex]

1. Show that [tex]\{ f_k \}[/tex] has a pointwise limit, [tex]f[/tex], but that

[tex]f_k \nrightarrow f[/tex] uniformly.

2.Does [tex]\displaystyle \int^1_0 f_k(x) dx \to \int^1_0 f(x) dx[/tex]?

2. Relevant equations

3. The attempt at a solution

1. Does [tex]\{ f_k \}[/tex] converges pointwise to [tex]f = 0[/tex] because for every [tex]x[/tex] there

is always a [tex]k[/tex] such that [tex]\displaystyle \frac{1}{k} < x[/tex]? But taking limit of each interval results in the following function

[tex]

\begin{align*}

\lim_{k \to \infty} f_k(x) = \left\{

\begin{array}{ll}

0 & 0 \\

\infty & \displaystyle 0 < x \leq \frac{1}{2k} \\

\infty & \displaystyle \frac{1}{2k} < x < \frac{1}{k} \\

0 & \displaystyle \frac{1}{k} \\

0 & \displaystyle \frac{1}{k} < x \leq 1

\end{array}

\right.

\end{align*}

[/tex]

Which one is the correct limit function? What is the reason for the sequence to be not uniformly convergent? Is it because the limit function [tex]f[/tex] broken into intervals is not continuous? But what if the limit function is [tex]f(x) = 0[/tex]?

2. Is this because limit and integrals cannot be exchanged due to sequence not uniformly convergent? What else do I need to prove?

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# Homework Help: Uniform convergence of function sequence

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