\(L^p\) convergence does not imply pointwise convergence or even a.e. convergence – instead, if \(f_k\to f\) in \(L^p\), there is some subsequence that converges to \(f\) a.e.
The following are dense subspaces of \(L^2([0, 1])\) and \(L^1({\mathbf{R}})\):
- Bounded measurable functions with bounded support
- Simple functions
- Step functions
- \(C_0([0, 1])\)
- Smoothly differentiable functions \(C_0^\infty([0, 1])\)
- Smooth compactly supported functions \(C_c^\infty\)
\begin{align*} m(X) < \infty \implies \lim_{p\to\infty} {\left\lVert {f} \right\rVert}_p = {\left\lVert {f} \right\rVert}_\infty .\end{align*}
Let \(M = {\left\lVert {f} \right\rVert}_\infty\).
- For any \(L < M\), let \(S = \left\{{{\left\lvert {f} \right\rvert} \geq L}\right\}\).
- Then \(m(S) > 0\) and
\begin{align*} {\left\lVert {f} \right\rVert}_{p} &= \left( \int_X {\left\lvert {f} \right\rvert}^p \right)^{\frac 1 p} \\ &\geq \left( \int_S {\left\lvert {f} \right\rvert}^p \right)^{\frac 1 p} \\ &\geq L ~m(S)^{\frac 1 p} \overset{p\to\infty}\to L \\ &\implies \liminf_p {\left\lVert {f} \right\rVert}_{p} \geq M .\end{align*}
We also have \begin{align*} {\left\lVert {f} \right\rVert}_{p} &= \left( \int_X {\left\lvert {f} \right\rvert}^p \right)^{\frac 1 p} \\ &\leq \left( \int_X M^p \right)^{\frac 1 p} \\ &= M ~m(X)^{\frac 1 p} \xrightarrow{p\to\infty} M \\ &\implies \limsup_p {\left\lVert {f} \right\rVert}_{p} \leq M .\end{align*}
For \(1\leq p< \infty\), \((L^p) {}^{ \vee }\cong L^q\).
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\todo[inline]{todo}
Use Riesz Representation for Hilbert spaces.
\(L^1 \subset (L^\infty) {}^{ \vee }\), since the isometric mapping is always injective, but never surjective.
\begin{align*} f_k \overset{{ \text{a.e.} }}\to f \text{ and } {\left\lVert {f_k} \right\rVert}_p \leq M \implies f\in L^p \text{ and } {\left\lVert {f} \right\rVert}_p \leq M .\end{align*}
Proof: Apply Fatou to \({\left\lvert {f} \right\rvert}^p\): \begin{align*} \int {\left\lvert {f} \right\rvert}^p = \int \liminf {\left\lvert {f_k} \right\rvert}^p \leq \liminf \int {\left\lvert {f_k} \right\rvert}^p = M .\end{align*}
\begin{align*} f \text{ uniformly continuous }: \quad {\left\lVert {\tau_h f - f} \right\rVert}_{L^p(X)} \overset{h\to 0}\longrightarrow 0 && \forall p .\end{align*}
Take \(g_k \in C_c^0 \to f\), then \(g\) is uniformly continuous, so \begin{align*} {\left\lVert {\tau_h f - f} \right\rVert}_p \leq {\left\lVert {\tau_h f - \tau_h g} \right\rVert}_p + {\left\lVert {\tau_h g - g} \right\rVert}_p + {\left\lVert {g - f} \right\rVert}_p \to 0 .\end{align*}
\begin{align*} (f, g) \in L^p\times L^q \implies f\ast g \text{ uniformly continuous} .\end{align*}
Use Young’s inequality \begin{align*} {\left\lVert {\tau_h(f\ast g) - f\ast g} \right\rVert}_\infty &= {\left\lVert {(\tau_h f - f) \ast g} \right\rVert}_\infty \leq {\left\lVert {\tau_hf - f} \right\rVert}_p {\left\lVert {g} \right\rVert}_q \to 0 .\end{align*}