[proofplan] We prove that the integral defining $\Lambda_g$ is meaningful on equivalence classes and then estimate it by Hölder's inequality. The only analytic input is Hölder's inequality in the full conjugate-exponent range, including the endpoint pairs $(1,\infty)$ and $(\infty,1)$. Once the estimate is established for every $f \in L^p$, linearity follows from linearity of the integral, and the operator-norm bound follows by taking the supremum over the unit ball of $L^p$. [/proofplan]

[step:Use Hölder's inequality to make the pairing finite]Fix $g \in L^q(X,\mathcal{A},\mu;\mathbb{F})$. Let $f \in L^p(X,\mathcal{A},\mu;\mathbb{F})$, and choose measurable representatives \begin{align*} \tilde f: X \to \mathbb{F} \end{align*} and \begin{align*} \tilde g: X \to \mathbb{F} \end{align*} of the equivalence classes $f$ and $g$, respectively. The pointwise product \begin{align*} h: X \to \mathbb{F}, \quad x \mapsto \tilde f(x)\overline{\tilde g(x)} \end{align*} is $\mathcal{A}$-measurable because it is obtained from measurable scalar-valued functions by multiplication and complex conjugation. By Hölder's inequality for conjugate exponents, including the endpoint cases $p=1,q=\infty$ and $p=\infty,q=1$ (citing a result not yet in the wiki: Hölder's inequality), applied to the [measurable functions](/page/Measurable%20Functions) $|\tilde f|$ and $|\tilde g|$, we have \begin{align*} \int_X |\tilde f(x)\overline{\tilde g(x)}|\,d\mu(x) \leq \|\tilde f\|_{L^p}\|\tilde g\|_{L^q}. \end{align*} Since $\tilde f \in L^p(X,\mathcal{A},\mu;\mathbb{F})$ and $\tilde g \in L^q(X,\mathcal{A},\mu;\mathbb{F})$, the right-hand side is finite. Hence $h \in L^1(X,\mathcal{A},\mu;\mathbb{F})$, so the integral \begin{align*} \int_X \tilde f(x)\overline{\tilde g(x)}\,d\mu(x) \end{align*} is finite and well-defined as a scalar integral.[/step]

[guided]Fix $g \in L^q(X,\mathcal{A},\mu;\mathbb{F})$. Because elements of $L^p$ and $L^q$ are equivalence classes of measurable functions, we begin with representatives. Let \begin{align*} \tilde f: X \to \mathbb{F} \end{align*} be a measurable representative of an arbitrary class $f \in L^p(X,\mathcal{A},\mu;\mathbb{F})$, and let \begin{align*} \tilde g: X \to \mathbb{F} \end{align*} be a measurable representative of the fixed class $g \in L^q(X,\mathcal{A},\mu;\mathbb{F})$. Define the product function \begin{align*} h: X \to \mathbb{F}, \quad x \mapsto \tilde f(x)\overline{\tilde g(x)}. \end{align*} This function is $\mathcal{A}$-measurable because $\tilde f$ and $\tilde g$ are measurable, complex conjugation is a measurable scalar operation, and multiplication is a measurable scalar operation. The point of Hölder's inequality here is twofold: it proves integrability of the product and gives exactly the norm estimate needed for boundedness. We apply Hölder's inequality for conjugate exponents to the non-negative measurable functions $|\tilde f|$ and $|\tilde g|$. The hypotheses are satisfied because $\tilde f \in L^p(X,\mathcal{A},\mu;\mathbb{F})$, $\tilde g \in L^q(X,\mathcal{A},\mu;\mathbb{F})$, and $p,q$ are conjugate exponents, including the endpoint possibilities $p=1,q=\infty$ and $p=\infty,q=1$. Thus Hölder's inequality gives \begin{align*} \int_X |\tilde f(x)||\tilde g(x)|\,d\mu(x) \leq \|\tilde f\|_{L^p}\|\tilde g\|_{L^q}. \end{align*} Since $|\tilde f(x)\overline{\tilde g(x)}|=|\tilde f(x)||\tilde g(x)|$ for every $x \in X$, this is the same as \begin{align*} \int_X |\tilde f(x)\overline{\tilde g(x)}|\,d\mu(x) \leq \|\tilde f\|_{L^p}\|\tilde g\|_{L^q}. \end{align*} The right-hand side is finite by the assumptions $\tilde f \in L^p$ and $\tilde g \in L^q$. Therefore $\tilde f\,\overline{\tilde g} \in L^1(X,\mathcal{A},\mu;\mathbb{F})$, and the scalar integral \begin{align*} \int_X \tilde f(x)\overline{\tilde g(x)}\,d\mu(x) \end{align*} is finite.[/guided]

[step:Check that the integral depends only on equivalence classes] Let $\tilde f_1:X\to\mathbb{F}$ and $\tilde f_2:X\to\mathbb{F}$ be measurable representatives of the same element $f \in L^p(X,\mathcal{A},\mu;\mathbb{F})$, and let $\tilde g_1:X\to\mathbb{F}$ and $\tilde g_2:X\to\mathbb{F}$ be measurable representatives of the same element $g \in L^q(X,\mathcal{A},\mu;\mathbb{F})$. Then $\tilde f_1=\tilde f_2$ $\mu$-a.e. and $\tilde g_1=\tilde g_2$ $\mu$-a.e. Hence \begin{align*} \tilde f_1\overline{\tilde g_1}=\tilde f_2\overline{\tilde g_2} \end{align*} $\mu$-a.e. Since both products are integrable by the previous step, equality almost everywhere of integrable functions gives \begin{align*} \int_X \tilde f_1(x)\overline{\tilde g_1(x)}\,d\mu(x)=\int_X \tilde f_2(x)\overline{\tilde g_2(x)}\,d\mu(x). \end{align*} Therefore the definition \begin{align*} \Lambda_g(f):=\int_X f\overline{g}\,d\mu(x) \end{align*} is independent of all choices of representatives. [/step]

[step:Verify linearity in the first argument] Let $f_1,f_2 \in L^p(X,\mathcal{A},\mu;\mathbb{F})$, and let $a,b \in \mathbb{F}$. Choose measurable representatives \begin{align*} \tilde f_1:X\to\mathbb{F}, \quad \tilde f_2:X\to\mathbb{F}, \quad \tilde g:X\to\mathbb{F} \end{align*} of $f_1$, $f_2$, and $g$, respectively. Then $a\tilde f_1+b\tilde f_2$ is a measurable representative of $af_1+bf_2$. By linearity of the scalar integral in the integrand and because $\tilde g$ is fixed, we obtain \begin{align*} \Lambda_g(af_1+bf_2)=\int_X (a\tilde f_1(x)+b\tilde f_2(x))\overline{\tilde g(x)}\,d\mu(x). \end{align*} Expanding the integrand and using linearity of the integral gives \begin{align*} \Lambda_g(af_1+bf_2)=a\int_X \tilde f_1(x)\overline{\tilde g(x)}\,d\mu(x)+b\int_X \tilde f_2(x)\overline{\tilde g(x)}\,d\mu(x). \end{align*} Thus \begin{align*} \Lambda_g(af_1+bf_2)=a\Lambda_g(f_1)+b\Lambda_g(f_2). \end{align*} So $\Lambda_g$ is linear over $\mathbb{F}$ in the first argument. In the real case, complex conjugation is the identity, and the same computation applies. [/step]

[step:Take the supremum over the unit ball to obtain the operator norm bound] For every $f \in L^p(X,\mathcal{A},\mu;\mathbb{F})$, the Hölder estimate from the first step gives \begin{align*} |\Lambda_g(f)| \leq \|f\|_{L^p}\|g\|_{L^q}. \end{align*} In particular, if $\|f\|_{L^p}\leq 1$, then \begin{align*} |\Lambda_g(f)| \leq \|g\|_{L^q}. \end{align*} By the definition of the [operator norm](/page/Operator%20Norm) of a linear functional on the normed space $L^p(X,\mathcal{A},\mu;\mathbb{F})$, \begin{align*} \|\Lambda_g\|_{(L^p)^*}=\sup\{|\Lambda_g(f)|: f\in L^p(X,\mathcal{A},\mu;\mathbb{F}), \|f\|_{L^p}\leq 1\}. \end{align*} Taking the supremum in the preceding estimate yields \begin{align*} \|\Lambda_g\|_{(L^p)^*}\leq \|g\|_{L^q}. \end{align*} Hence $\Lambda_g$ is bounded, and the asserted norm estimate follows. [/step]