Counterexamples

Non-integrable functions

example (Examples of integrable functions):

$\int {1\over 1 + x^2} = \arctan(x) \overset{x\to\infty}\to \pi/2 < \infty$
Any bounded function (or continuous on a compact set, by EVT)
$\int_0^1 {1 \over \sqrt{x}} < \infty$
$\int_0^1 {1\over x^{1-{\varepsilon}}} < \infty$
$\int_1^\infty {1\over x^{1+{\varepsilon}}} < \infty$

example (Examples of non-integrable functions):

$\int_0^1 {1\over x} = \infty$ .
$\int_1^\infty {1\over x} = \infty$ .
$\int_1^\infty {1 \over \sqrt{x}} = \infty$
$\int_1^\infty {1\over x^{1-{\varepsilon}}} = \infty$
$\int_0^1 {1\over x^{1+{\varepsilon}}} = \infty$

proposition (a.e. convergence never implies

$L^p$ convergence):

Sequences $f_k \overset{a.e.}\to f$ but $f_k \overset{L^p}{\not\to} f$ :

For $1\leq p < \infty$ : The skateboard to infinity, $f_k = \chi_{[k, k+1]}$ .

Then $f_k \overset{a.e.}\to 0$ but ${\left\lVert {f_k} \right\rVert}_p = 1$ for all $k$ .

Converges pointwise and a.e., but not uniformly and not in norm.
For $p = \infty$ : The sliding boxes $f_k = k \cdot \chi_{[0, \frac 1 k]}$ .

Then similarly $f_k \overset{a.e.}\to 0$ , but ${\left\lVert {f_k} \right\rVert}_p = 1$ and ${\left\lVert {f_k} \right\rVert}_\infty = k \to \infty$

Converges a.e., but not uniformly, not pointwise, and not in norm.

remark:

Notions of convergence:

Uniform
Pointwise
Almost everywhere
In norm

Uniform $\implies$ pointwise $\implies$ almost everywhere, but in general non of these can be reversed.

proposition (The four big counterexamples in convergence):

Uniform: $f_n \rightrightarrows f: \forall \varepsilon ~\exists N {~\mathrel{\Big\vert}~}~n\geq N \implies {\left\lvert {f_N(x) - f(x)} \right\rvert} < \varepsilon \quad \forall x.$
Pointwise: $f_n(x) \to f(x)$ for all $x$ . (This is just a sequence of numbers)
Almost Everywhere: $f_n(x) \to f(x)$ for almost all $x$ .
Norm: ${\left\lVert {f_n - f} \right\rVert}_1 = \int {\left\lvert {f_n(x) - f(x)} \right\rvert} \to 0$ .

We have $1 \implies 2 \implies 3$ , and in general no implication can be reversed, but (warning) none of $1,2,3$ imply $4$ or vice versa.

$f_n = (1/n) \chi_{(0, n)}$ . This converges uniformly to 0, but the integral is identically 1. So this satisfies 1,2,3 and not 4.
$f_n = \chi_{(n, n+1)}$ (skateboard to infinity). This satisfies 2,3 but not 1, 4.
$f_n = n\chi_{(0, \frac 1 n)}$ . This satisfies 3 but not 1,2,4.
$f_n:$ one can construct a sequence where $f_n \to 0$ in $L^1$ but is not 1,2, or 3. The construction:
- Break $I$ into $2$ intervals, let $f_1$ be the indicator on the first half, $f_2$ the indicator on the second.
- Break $I$ into $2^2=4$ intervals, like $f_3$ be the indicator on the first quarter, $f_4$ on the second, etc.
- Break $I$ into $2^k$ intervals and cyclic through $k$ indicator functions.
- Then $\int f_n = 1/2^n \to 0$ , but $f_n\not\to 0$ pointwise since for every $x$ , there are infinitely many $n$ for which $f_n(x) = 0$ and infinitely many for which $f_n(x) = 1$ .

Almost everywhere convergence does not imply $L^p$ convergence for any $1\leq p \leq \infty$ . In the following examples, $f_k \overset{a.e.}\to f$ but $f_k \overset{L^p}{\not\to} f$ :

For $1\leq p < \infty$ : The skateboard to infinity, $f_k = \chi_{[k, k+1]}$ .
- Then $f_k \overset{a.e.}\to 0$ but ${\left\lVert {f_k} \right\rVert}_p = 1$ for all $k$ .
- Converges pointwise and a.e., but not uniformly and not in norm.
For $p = \infty$ : The sliding boxes $f_k = k \cdot \chi_{[0, \frac 1 k]}$ .
- Then similarly $f_k \overset{a.e.}\to 0$ , but ${\left\lVert {f_k} \right\rVert}_p = 1$ and ${\left\lVert {f_k} \right\rVert}_\infty = k \to \infty$
- Converges a.e., but not uniformly, not pointwise, and not in norm.

proposition (Functional analytic properties of

$L^1$ and

$L^2$ ):

For any measure space $(X, {\mathcal{M}}, \mu)$ ,

$L^1(X)$ is Banach space.
$L^2(X)$ is a (possibly non-separable) Hilbert space.