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Is the subspace $\{f \in L^\infty(\mathbb{R}) ~|~ f \text{ is continuous at } x=0 \}$ a Banach space? The norm is of course the essential supremum.

Does the essential supremum even notice a single continuity point? Is there a simple argument to prove or contradict the closeness of this subspace?

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Let us assume that $f_n \to f$ in $L^\infty(\mathbb{R})$ and that all $f_n$ are continuous in $0$.

In a first step, show that $f_n(0)$ converges. This can be easily seen since $|f_n(0) - f_m(0)| \le \|f_n - f_m\|_\infty$, which follows from the continuity of $f_n$ and $f_m$. We call the limit $f(0)$.

Let us show that $f$ is continuous in $0$ by using an $\varepsilon$-$\delta$ argument. So let $\varepsilon > 0$ be given. Then, there is $n$, such that $\|f_n - f\|_\infty \le \varepsilon$. And there is $\delta > 0$, such that $|f_n(x) - f_n(0)| \le \varepsilon$ for all $x$ with $|x|\le \delta$. Now we have $$|f(0) - f(x)| \le |f(0) - f_n(0)| + |f_n(0) - f_n(x)| + |f_n(x) - f(0)| \le 3 \, \varepsilon$$ for almost all $x$ with $|x|\le \delta$. Hence, $f$ is continuous in $0$.

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    $\begingroup$ It is not completely clear that $|f_n(0) - f_m (0)| \leq \| f_n - f_m\|_\infty$, since $\|\cdot\|_\infty$ is onle the essential supremum, not the supremum. You will have to use continuity of the $f_n$ to show this. $\endgroup$ – PhoemueX Dec 13 '15 at 19:51
  • $\begingroup$ Yes, this follows from the continuity of $f_n$. I should have mentioned this. $\endgroup$ – gerw Dec 13 '15 at 19:52
  • $\begingroup$ To give a rigorous argument: Let $f$ be continuous at $0$ and let $\varepsilon >0$ be arbitrary. There is $\delta > 0$ with $|f (x) - f (0)| < \varepsilon$ for $|x|<\delta$. Since $(\delta, \delta)$ is not a null set, we get $\|f\|_\infty \geq \inf_{|x|<\delta} |f(y)| \geq |f(0)| - \varepsilon$. Since $\varepsilon > 0$ was arbitrary, $\|f\|_\infty \geq |f(0)|$. $\endgroup$ – PhoemueX Dec 13 '15 at 19:56
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The answer is yes. I'll deal only with the completeness question, and sketch the proof of a special case: Suppose $f_n$ is a sequence of actual bounded functions defined everywhere, with $f_n(0)=$ for all $n$ and each $f_n$ continuous at $0.$ Suppose further $[f_n]\to [f]$ in $L^\infty.$ (Here $[\,]$ denotes the familiar $L^\infty $ equivalence class.) Then there are sets $E_n$ of measure $0$ such that

$$\sup_{\mathbb R\setminus E_n} |f-f_n| = \|f-f_n\|_\infty.$$

These are actual functions here. Set $E = \mathbb R\setminus \cup E_n.$ Then on $\mathbb R\setminus E,$ we have $f_n \to f$ uniformly. Hence we can use the classical result on interchanging limits in the presence of uniform convergence: As $x\to 0$ within $\mathbb R \setminus E,$

$$\lim_{x\to 0} \lim_{n\to \infty} f_n(x) = \lim_{n\to \infty}\lim_{x \to 0} f_n(x) = 0.$$

That gives the result.

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