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In Evan's book "Partial Differential Equations", it says that:

If $u: U \to \mathbb{R}$ is bounded and continuous, we write $$\|u \|_{C(\overline U)} = \sup_{x\in U} |u(x)| $$

But, the notation $\| \cdot \|_{C(\overline U)} $ seems to imply that the function $u$ is continuous up to the boundary, so is it correct that the function $u$ which is continuous and bounded can be extended continuously up to the boundary?

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    No, but what is $U$? – 2017-01-02
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    Of course not: a classic counterexample is $\sin (1/x)$ defined on $(0,1)$. In any case, the supremum is defined without requiring continuity at the boundary. – 2017-01-02
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    Yes, its true that can be extended – 2017-01-02
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    @JonasMeyer $U$ is open bounded subset in $R^n$ – 2017-01-02
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    No, as @Crostul says, such functions need not extend continuously to the closure. It is indeed strange notation. – 2017-01-02

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No, as Crostul commented, even for $U=(0,1)$ in $\mathbb R$, as the example $\sin(1/x)$ shows there.

For the continuous extension to exist, it is necessary and sufficient that the function is uniformly continuous. "Necessary" because $\overline U$ is compact. "Sufficient" because a uniformly continuous function from a metric space to $\mathbb R$ can be uniquely continuously extended to the completion.

As zhw commented, this is strange notation.

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    The notation is not strange. Evans defines the space $C(\bar{U})$ as the space of uniformly continuous real-valued functions on $U$. The notation is meant to remind you that you can extend such functions to the boundary continuously. – 2017-01-02
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    @Jeff But, if you know onl that the function $u$ is bounded and continuous, then it is not necessarily uniformly continuous, so the notation is indeed strange. Isn't it? – 2017-01-02
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    He's just defining the norm on the space $C(\bar{U})$. Functions in this space are by definition uniformly continuous. This is standard notation in PDEs. – 2017-01-02
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    @Jeff The function $u$ being bounded and continuous is not necessarily in that space though – 2017-01-02
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    Sure, so the norm applies to a larger class of functions (in fact, continuity is not necessary either). It's very common to denote a norm on a Banach space as $\|u\|_X$ where $X$ is the name of the space. That's all that is going on here. – 2017-01-02