Uniform convergence of function sequence

[complexnumber]

Homework Statement

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

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 $$\displaystyle \frac{1}{k} < x$$? But taking limit of each interval results in the following function
$$\begin{align*}<br /> \lim_{k \to \infty} f_k(x) = \left\{<br /> \begin{array}{ll}<br /> 0 & 0 \\<br /> \infty & \displaystyle 0 < x \leq \frac{1}{2k} \\<br /> \infty & \displaystyle \frac{1}{2k} < x < \frac{1}{k} \\<br /> 0 & \displaystyle \frac{1}{k} \\<br /> 0 & \displaystyle \frac{1}{k} < x \leq 1<br /> \end{array}<br /> \right.<br /> \end{align*}$$

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 $\frac{1}{k} < x_0$, 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:$\forall \epsilon > 0,$ there exists a K such that k > K implies $|f(x) - f_k(x)| < \epsilon$ $\forall x \in \left[ 0,1 \right]$. 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

$$\displaystyle \int^1_0 f_k(x) dx$$