Consider an open interval $(0,1) \subset \mathbb{R}$ and the subset $$ \mathcal{F} := \{f \in L^1((0,1)): \|{f\vert_{(a,b)}}\|_{\infty} = \infty \, \forall \, 0 \leq a < b < 1\} \subset L((0,1), dx). $$ I want to show that $\mathcal{F}$ is non-empty in $(L((0,1), dx), \| \, \|_{L^1})$.
For that reason, I define the sets $$G_n := \{f \in L^1((0,1)): \|{f\vert_{(a,b)}}\|_{\infty} \leq n \, \text{ for some } 0 \leq a < b < 1\}$$ and aim to show two things:
- $G_n$ is closed for every $n \in \mathbb{N}$
- $G_n$ has empty interior for $n \in \mathbb{N}$
Then, $F_n := L^1{((0,1))}\setminus G_n$ is open and dense for every $n \in \mathbb{N}$ and $$ \mathcal{F} = \bigcap_{n \geq 1} F_n$$ is non-empty by Baire's category theorem (actually, it will be comeager).
Now, closedness of $(G_n; n \geq 1)$ is clear.
To prove that $G_n$ have empty interior for every $n$, I assume, for the sake of contradiction, that they have not. That is, let $n \in \mathbb{N}$. Then, for every $g \in G_n$, there exists $\varepsilon > 0$ s.t.
$$ B := \{h \in L^{1}((0,1)): \|h - g \|_{L^1} < \varepsilon\} \subset G_n.$$ The idea is now to construct a function $h \in L^1((0,1))$ s.t. $h \in B$ and $h \notin G_n$, i.e. $$ \|h-g\|_{L^1} < \varepsilon /2 \quad \text{ and} \quad \|h\vert_{(a,b)} \|_{\infty} > n \quad \forall \, 0 < a < b < 1.$$ Is this a feasible approach? If yes, can you hint how this may work?
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ferhenk
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