There’s a lot of examples of pointwise but not uniform convergence. But I suppose it wouldn’t harm to have another one.

## Motivation

Imagine a wave packet that travels horizontally from the origin to infinity with a velocity of $$1$$. Let $$f_t(x)$$ be a snapshot of that wave at time $$t$$

It’s clear that as time progresses, the wave packet would eventually pass through all points in $$\mathbb{R}^+$$.

Focusing on one point $$x_0$$, once the wave passes through it completely, the amplitude on that point would be $$0$$ from then˚ on.

However for all time $$t$$, there’s always a point on $$\mathbb{R}^+$$ such that the amplitude is nonzero.

## Construction

Let $$f_t(x)$$ be a sequence of functions from $$\mathbb{R}$$ to $$\mathbb{R}$$.

$f_t(x) := \left\{ \begin{array}{ll} x-(t-1) & \text{if } t-1 \leq x \leq t \\ -(x-(t+1)) & \text{if } t < x \leq t+1 \\ 0 & \text{otherwise} \end{array} \right.$ Clearly $$f_t(x)$$ is continuos on $$\mathbb{R}$$.

Let $$x_0$$ be some real number. If $$t > x_0 + 1$$, then $$f_t(x_0) = 0$$. So $$\lim_{t \to \infty} f_t(x_0) = 0$$ and hence $$(f_t(x))$$ is pointwise convergent to $$0$$.

However, $$f_t(t) = 1$$ for all $$t \in \mathbb{R}$$, so $$\lim_{t \to \infty} \sup \{\lvert f_t(x) - f(x) \rvert : x \in \mathbb{R}\} = 1 \neq 0$$ and hence $$(f_t(x))$$ does not uniformly converge to $$0$$.

## Fatou’s Lemma

It’s also a nice function to demonstrate Fatou’s lemma. We can observe

$0 = \int \liminf_{t\to \infty} f_t(x) \leq \liminf_{t\to \infty} \int f_t(x) = 1$