Uniform convergence of function sequence

[complexnumber]

Homework Statement

For k = 1,2,\ldots define f_k : [0,1] \to \mathbb{R} by
<br /> \begin{align*}<br /> f_k(x) = \left\{<br /> \begin{array}{ll}<br /> 4k^2x &amp; 0 \leq \displaystyle x \leq \frac{1}{2k} \\<br /> 4k(1 - kx) &amp; \displaystyle \frac{1}{2k} &lt; x \leq \frac{1}{k} \\<br /> 0 &amp; \displaystyle \frac{1}{k} &lt; x \leq 1<br /> \end{array}<br /> \right.<br /> \end{align*}<br />

1. Show that \{ f_k \} has a pointwise limit, f, but that
f_k \nrightarrow f uniformly.

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

Homework Equations

The Attempt at a Solution

1. Does \{ f_k \} converges pointwise to f = 0 because for every x there
is always a k such that \displaystyle \frac{1}{k} &lt; x? But taking limit of each interval results in the following function
<br /> \begin{align*}<br /> \lim_{k \to \infty} f_k(x) = \left\{<br /> \begin{array}{ll}<br /> 0 &amp; 0 \\<br /> \infty &amp; \displaystyle 0 &lt; x \leq \frac{1}{2k} \\<br /> \infty &amp; \displaystyle \frac{1}{2k} &lt; x &lt; \frac{1}{k} \\<br /> 0 &amp; \displaystyle \frac{1}{k} \\<br /> 0 &amp; \displaystyle \frac{1}{k} &lt; x \leq 1<br /> \end{array}<br /> \right.<br /> \end{align*}<br />

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 f broken into intervals is not continuous? But what if the limit function is f(x) = 0?

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

 

[Hoblitz]

I find it easier to think about these questions by picking arbitrary x_0 in the interval and seeing what happens. If you pick zero,f_k (0) = 0 and this is trivial. If x_0 is non-zero, indeed you can find large enough k such that \frac{1}{k} &lt; x_0, and for that k and larger,f_k (x_0) = 0. This is for arbitrary x_0, so you can see how the limit function is just f = 0.

The main problem for you seems to be determining the uniform convergence. I would work straight from the definition:\forall \epsilon &gt; 0, there exists a K such that k > K implies |f(x) - f_k(x)| &lt; \epsilon \forall x \in \left[ 0,1 \right]. In your case, f(x) = 0 for all x.

I think it's easiest to work from a contradiction. Suppose you had such an epsilon, and such a K. What would happen near zero?

As for the second part, I think it would be very informative to actually evaluate the integral

<br /> \displaystyle \int^1_0 f_k(x) dx<br />