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$$C([a,b])=\{ f: [a,b] \to \mathbb{R} \text{ continuous} \}$$

$C([a,b])$ is a linear space.

For $f \in C([a,b])$ we define $\|f\|_{\infty}:= \sup_{x \in [a,b]} |f(x)|$ and easily it can be shown that the space $(C([a,b]),\| \cdot\|_\infty)$ is a space with norm.

For $f \in C([a,b])$ we define $\|f\|:= \int_a^b |f(x)| \, dx$

and it can be easily shown that the space $(C([a,b]), \|\cdot\|)$ is a space with norm.

$$(f,g) \mapsto \|f-g\|=\int_a^b |f(x)-g(x)|\,dx$$

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There is a $f_n \in C([a,b])$ such that $\|f_n-g\| \to 0$ but $g \notin C([a,b])$.

Could you give me an example of such a function?

Also does the last proposition also hold if we consider $\|f\|_\infty:= \sup_{x \in [a,b]} |f(x)|$ ?

Michael Hardy
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evinda
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    What is $g$? Perhaps this is what you're after. – David Mitra May 21 '15 at 20:23
  • When you write $|f|$ instead of $||f||$ then you see a $|f||g|$ instead of $||f||||g||$, and you see the difference in spacing. The former notation is standard and I changed it in the question. ${}\qquad{}$ – Michael Hardy May 21 '15 at 20:25

2 Answers2

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Hint: Consider the set of functions

$$f_n(x) = \begin{cases} 0, & x \in \left[0, \frac{1}{2}-\frac{1}{2n}\right) \\ nx + \frac{1-n}{2}, & x \in \left[\frac{1}{2}-\frac{1}{2n},\frac{1}{2}+\frac{1}{2n}\right] \\ 1, & x\in\left(\frac{1}{2}+\frac{1}{2n},1\right]\end{cases}$$

Does this seemingly converge to anything in the $L^1$ norm? If so, is that function continuous? This can easily be adapted to $C([a,b])$ in general.

Cameron Williams
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Whenever $f_n\rightarrow g$ you can change a value of $g$ such that $g$ is not continuous. But changing a value of $g$ does not change $\int_a^b |f_n(x)-g(x)|dx$ so you still have $f_n\rightarrow g$...

Stephan Kulla
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