Same on dense subspace implies same on whole space?

Question

I read the whole proof of Theorem 13.18 Bruckner's Real Analysis book and I had no problem understanding the proof except for following two claims inside the proof that are stated without further explanations.

One is dense-ness of $L^{\infty}$ in $L^{p}$ which the post in here is not very clear. How $L^{\infty}$ is dense in $L^{p}$?

The second : because two functionals $\Gamma_1, \Gamma : L^{p} \to \mathbb{F}$ agree on a dense subspace of $L^{p}$ how that implies they agree on whole domain $L^{p}$?

I would appreciate a detailed guidance about the two questions.

Answer 1

Claim: Let $f,g: X \to Y$ be two continuous functions (between metric spaces) such that $f$ and $g$ agree on a dense subspace $D$, then $f=g$.

Proof: Let $x \in X$ and choose a sequence $\{d_n\}_n$ in $D$ with $\lim_n d_n = x$. Then $$f(x)= \lim_n f(d_n) = \lim_n g(d_n) = g(x)$$ so $f=g$.

For your other question, the answer in that question proves the stronger statement that $L^1 \cap L^\infty$ is dense in $L^p$, so certainly $L^\infty$ is dense in $L^p.$

Answer 2

Item 1: The Theorem 13.18 is about $\sigma$-finite measure spaces and $L^p$, where $1\leq p < \infty$. The proof begins by considering the case of finite measure spaces.

So, let $(X, \mathcal{M}, \mu)$ be a measure space and $\mu(X) <\infty$. Given any $g \in L^\infty$, we have that there is $M>0$ such that $|g| \leq M$ almost everywhere. So, given any $p$, $1\leq p < \infty$, we have: $$\int_X |g|^p d\mu \leq \int_X M^p d\mu = M^p \mu(X) < \infty$$ So, we have that $L^\infty \subseteq L^p$.

Now, given any $f \in L^p$, then, for all $n \in \Bbb N$, let $f_n= f \cdot \chi_{\{x \in X : |f(x)| \leq n\}}$. It is easy to see that $f_n \in L^\infty$.

Note that $| f - f_n|^p = |f|^p \cdot \chi_{\{x \in X : |f(x)| > n\}}$. Since $f \in L^p$, $f$ is finite almost everywhere, so, as $n \to \infty$, we have $|f|^p \cdot \chi_{\{x \in X : |f(x)| > n\}} \to 0$ almost everywhere. Note also that $|f|^p \cdot \chi_{\{x \in X : |f(x)| > n\}} \leq |f|^p$ and, since $f$ is in $L^p$,we have that $|f|^p \in L^1$. Then, applying the Dominated Convergence Theorem, we have $$ \lim_n \int_X | f - f_n|^p d\mu = \lim_n \int_X |f|^p \cdot \chi_{\{x \in X : |f(x)| > n\}} d\mu = \int_X 0 \ d\mu =0$$

Remark: After proving $L^\infty \subseteq L^p$, if you know that the simple functions (which are functions in $L^\infty$) are dense in $L^p$, you can use this fact to deduce that $L^\infty$ is dense in $L^p$.

Item 2: You ask: "because two functionals $\Gamma_1, \Gamma : L^{p} \to \mathbb{F}$ agree on a dense subspace of $L^{p}$ how that implies they agree on whole domain $L^{p}$?"

The answer is yes, because the functionals $\Gamma_1, \Gamma : L^{p} \to \mathbb{F}$ in Theorem 13.18 are continuous linear functionals.

In fact, as QuantumSpace noted, this is a general property:

Let $X$ and $Y$ be metric spaces and let $D$ be a dense subset of $X$. Then, for any two continuous functions $f: X \rightarrow Y$ and $g:X \rightarrow Y$, if $f$ and $g$ agrees on $D$ then $f=g$

Proof: Given any $x \in X$, let $\{x_n\}_n$ be a sequence of points in $D$ such that $x_n \to x$. Then $$ f(x) = \lim_n f(x_n) = \lim_n g(x_n) =g(x) $$