---- Throughout, $\mathbb{F}$ denotes $\mathbb{R}$ or $\mathbb{C}$. $(X, \Sigma, \mu)$ is a [[measure|measure space]], and $f:X \to \mathbb{F}$ is a [[measurable function]]. $p'$ is the [[dual exponent]] to $p$. > [!theorem] Theorem. ([[converse to Hölder's inequality]]) > An application of [[Hölder's inequality]] shows that [[Lp-norm|for]] $1 \leq p < \infty$, $\|f\|_{p}=\sup_{ \|h\|_{p'} \leq 1 : f \overline{h} \text{ integrable} } \left| \langle f, h \rangle \right|=\sup_{ \|h\|_{p'} \leq 1 : f \overline{h} \text{ integrable} } \left|\int f \overline{h} \, d\mu \right|.$ > For $p=\infty$ the same identity holds provided $\mu$ is [[finite measure|σ-finite]], i.e., $\|f\|_{\infty}= \sup_{ \|h\|_{p'} \leq 1 : f \overline{h} \text{ integrable} } |\langle f,h \rangle | =\sup_{\|h\|_{1} \leq 1 : f \overline{h} \text{ integrable}} \left|\int f \overline{h} \, d\mu \right|.$ > > (If $\|f\|_{p}=+\infty$, the right-hand side is $+\infty$ as well. I think we should assume something like $\sigma$-finiteness in such a case to make sure this works.) ^theorem > [!equivalence] > Since $h \mapsto \overline{h}$ is a [[bijection]] of the unit ball, an equivalent statement without [[complex conjugate|complex conjugation]] is $\|f\|_{p}= \sup_{\|h\|_{p'} \leq 1: fh \text{ integrable}} \left|\int fh \, d\mu \right|,$ holding for all $1 \leq p < \infty$ and for $p=\infty$ if $\mu$ is [[finite measure|σ-finite]]. ^equivalence > [!note] Remarks. > 1. If we already know $f \in L^{p}(\mu)$, then the stipulation of $f \overline{h}$ being integrable is automatically satisfied, for in this case $\int f \overline{h} \, d\mu \leq |\int f \overline{h} \, d\mu | \leq \int |f \overline{h}| \, d\mu \leq \underbrace{ \|f\|_{p} }_{ < \infty } \underbrace{ \|h\|_{p'} }_{ \leq 1 } \, d\mu <\infty$ (the last $\leq$ has used [[Hölder's inequality]]). >2. This result is essentially restating the [[operator norm|identity]] $\|f\|_{p}=\|\langle f,- \rangle\|_{p'}$ as claimed in the '[[Lipschitz continuous|isometry]]' part of [[Lp duality]] when the latter's assumptions are satisfied. Of course, we cannot *conclude* this result from [[Lp duality]], since we will use it to *prove* [[Lp duality]]. This result also holds in some more general settings than [[Lp duality]]. ^note > [!proof] Proof. > > Since the map $h \mapsto \overline{h}$ is a [[bijection]] of the unit ball, it suffices prove the result without worrying about [[complex conjugate|complex conjugates]] (see the equivalence). > > [[Hölder's inequality]] implies that if $h \in \mathcal{L}^{p'}(\mu)$ with $\|h\|_{p'} \leq 1$, then $\begin{align} > |\int fh \, d\mu | \leq \int |fh| \, d\mu \leq \|f\|_{p} \|h\|_{p'} \leq \|f\|_{p} > \end{align}$ > from which RHS $\leq$ LHS follows. We proceed to prove the reverse inequality. > > If $\|f\|_{p}=0$ [[dual exponent|then]] by the [[triangle inequality for integrals]] and [[Hölder's inequality]] $|\int fh \, d\mu| \leq \int |fh| \, d\mu \leq \|f\|_{p} \|h\|_{p'}=0$ > and so the result holds. So assume $\|f\|_{p} \neq 0$. > > ### Case 1: $0 < \|f\|_{p}<\infty.$ > > #### Case 1a: $1 \leq p < \infty.$ > It suffices to find $h \in \mathcal{L}^{p'}(\mu)$, $\|h\|_{p'} \leq 1$, such that $\|f\|_{p} \leq |\int fh \, d\mu|$. To this end, define[^4] $h(x)=\begin{cases} > \frac{\overline{f(x)}|f(x)|^{p-2}}{\|f\|_{p}^{p/p'}} & f(x) \neq 0 \\ > 0 & f(x)=0. > \end{cases}$ > Then $\|h\|_{p'} =1$[^1] and $\int fh \, d\mu=\|f\|_{p}$.[^2] > > #### Case 1b: $p=\infty$ (now assuming $\mu$ is $\sigma$-finite) > Using the assumption that $\mu$ is a $\sigma$-[[finite measure|finite]] [[measure]], pick measurable sets $(X_{N})_{N \in \mathbb{N}}$ such that $\bigcup_{N=1}^{\infty} X_{N}=X$. Then there exists $N$ such that $0<\mu(X_{N}) < \infty$. For $\varepsilon>0$, let $X_{\varepsilon, N}:= \{ x \in X: |f(x)| \geq \|f\|_{\infty} - \varepsilon \} \cap X_{N}.$ > The set $X_{\varepsilon, N}$ has positive [[measure]] (otherwise $\|f\|_{\infty}$ would not be the [[Lp-norm|essential supremum]]). [[characteristic function|Define]] $h_{0}(x):= \begin{cases} > \frac{1}{\mu(E_{\varepsilon, N})} \frac{\overline{f(x)}}{|f(x)|} \chi_{E_{\varepsilon, N}}(x) & f(x) \neq 0 \\ > 0 & f(x)=0. > \end{cases}$ > Then $\|h_{0}\|_{1}=1$ and $\int fh_{0} \, d\mu = \frac{1}{\mu(E_{\varepsilon, N})} \int_{x \in E_{\varepsilon, N}} \overbrace{ |f(x)| }^{ \geq \|f\|_{\infty}-\varepsilon }\, d\mu(x) \geq \|f\|_{\infty}-\varepsilon.$ > Sending $\varepsilon \to 0$ obtains the desired inequality.[^5] > > > [^5]: To be pedantic, we are using the following fact: if $a,b \in \mathbb{R}$ are such that $a \geq b-\varepsilon$ for all $\varepsilon>0$, then $a \geq b$. Indeed, contrapositively suppose $a<b$. Put $\varepsilon:=\frac{b-a}{2}$. Then $a<b-\varepsilon=\frac{a+b}{2}$. > > > ### Case 2: $\|f\|_{p}=\infty.$ > The idea (a common one in analysis) is to truncate ourselves into the finite case $1$ and then $\sup$ over the truncations. > > Let $\alpha$ denote the [[supremum]] on the RHS of the desired equality. We have to show $\alpha=\infty$. We're going to do this for the [[Lebesgue measure]] on $\mathbb{R}^{m}$ with base [[field]] $\mathbb{R}$ because that's what is examinable. (I think in general something like $f_{k}(x):=\min\begin{cases}|f(x)|, k \}\chi_{X_{k}}(x) & (k \in \mathbb{N})\end{cases}$ would work in place of the $\mathbb{R}$-specific construction below. > > Define the truncations $f_{k}(x):=\begin{cases} > f(x) & \|x\| \leq k \text{ and } |f(x)| \leq k \text{ and } \\ > 0 & \text{else.} > \end{cases}$ > Clearly $f_{k} \to f$ [[pointwise converge|pointwise]]. > > ![[Pasted image 20251017152614.png|500]] > > Clearly $f_{k} \in L^{p}(X)$, [[integral|since it is bounded and of finite support]]. By the [[monotone convergence theorem for nonnegative measurable functions|monotone convergence theorem]] (and I think implicitly also [[the sequential continuity lemma|sequential continuity]] [[continuous|of]] $|\cdot|$ and $(\cdot)^{p}$ to factor the limit through the absolute value and exponentiation) > > $\lim_{k \to \infty}\|f_{k}\|^{p}_{p}= \lim_{k \to \infty}\int |f_{k}|^{p} \, d\mu = \int |f|^{p} \, d\mu $ > which equals $\infty$ by assumption. Thus $\lim_{k \to \infty} \|f_{k}\|_{p}=\infty$. > > > Case $1$ supplies, for each $k$, a function $h_{k}$, $\|h_{k}\|_{p'}=1$, for which $\|f_{k}\|_{p}^{} \leq |\int fh_{k} \, d\mu|$, from which it follows that $|\int fh_{k} \, d\mu | \xrightarrow[k \to \infty]{}\infty.$ > Thus $\alpha=\infty$, as desired. > [^1]: Verification: TODO bring over from written notes. [^2]: Verification: using $f(x) \overline{f(x)}=|f(x)|^{2}$, $\begin{align} \int fh \, d\mu &= \int_{x \in X} \frac{f(x) \overline{f(x)} |f(x)|^{p-2}}{\|f\|_{p}^{p/p'}} \, d\mu \\ &= \frac{1}{\|f\|_{p}^{p/p'}} \int _{x \in X} {|f(x)|^{p}}{} \, d\mu \\ &= \frac{\|f\|_{p}^{p}}{\|f\|_{p}^{p/p'}}\\ &= \|f\|_{_{p}}, \end{align}$ where the last step [[dual exponent|used that]] $p-\frac{p}{p'}=1$. ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```