Non-integrable functions
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\(\int {1\over 1 + x^2} = \arctan(x) \overset{x\to\infty}\to \pi/2 < \infty\)
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Any bounded function (or continuous on a compact set, by EVT)
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\(\int_0^1 {1 \over \sqrt{x}} < \infty\)
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\(\int_0^1 {1\over x^{1-{\varepsilon}}} < \infty\)
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\(\int_1^\infty {1\over x^{1+{\varepsilon}}} < \infty\)
- \(\int_0^1 {1\over x} = \infty\).
- \(\int_1^\infty {1\over x} = \infty\).
- \(\int_1^\infty {1 \over \sqrt{x}} = \infty\)
- \(\int_1^\infty {1\over x^{1-{\varepsilon}}} = \infty\)
- \(\int_0^1 {1\over x^{1+{\varepsilon}}} = \infty\)
Sequences \(f_k \overset{a.e.}\to f\) but \(f_k \overset{L^p}{\not\to} f\):
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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.
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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.
Notions of convergence:
- Uniform
- Pointwise
- Almost everywhere
- In norm
Uniform \(\implies\) pointwise \(\implies\) almost everywhere, but in general non of these can be reversed.
- 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.
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\(f_n = (1/n) \chi_{(0, n)}\). This converges uniformly to 0, but the integral is identically 1. So this satisfies 1,2,3 and not 4.
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\(f_n = \chi_{(n, n+1)}\) (skateboard to infinity). This satisfies 2,3 but not 1, 4.
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\(f_n = n\chi_{(0, \frac 1 n)}\). This satisfies 3 but not 1,2,4.
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\(f_n:\) 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 \(f_k \overset{L^p}{\not\to} f\):
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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.
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For \(p = \infty\): The sliding boxes \(f_k = k \cdot \chi_{[0, \frac 1 k]}\).
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Then similarly \(f_k \overset{a.e.}\to 0\), but \({\left\lVert {f_k} \right\rVert}_p = 1\) and \({\left\lVert {f_k} \right\rVert}_\infty = k \to \infty\)
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Converges a.e., but not uniformly, not pointwise, and not in norm.
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For any measure space \((X, {\mathcal{M}}, \mu)\),
- \(L^1(X)\) is Banach space.
- \(L^2(X)\) is a (possibly non-separable) Hilbert space.