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Show that $C_0[a,b]:= \{f \in C[a,b] : f(a)=f(b)=0 \}$ is $\| \cdot \|_2$-dense in $C[a,b]$. Is it also $\| \cdot \|_{\infty}$-dense in $C[a,b]$?

In the relevant chapter of my textbook, I am given the following definition:

Let $(\Omega,d)$ be a metric space, and let $A \subset \Omega$. The closure of $A$ (in $\Omega$) is the set

$\overline{A} := \{ x \in \Omega :$ there is a sequence $(a_n)_{n \in \mathbb{N}}$ in $A$ with $a_n \to x$}.
The set $A$ is called dense (in $\Omega$) if $\overline{A} = \Omega$.

Furthermore, since we are working in the space of continuous functions, I know that the associated $\| \cdot \|_2$ norm is defined as $\displaystyle \left( \int_a^b |f|^2 dx \right)^{\frac{1}{2}}$ and the $\| \cdot \|_{\infty}$ norm is defined as
$\sup \{|f(t)| : t \in [a,b]\}$.

Based on what I have read, one way to interpret a set being dense is if all the points, in this case, in $C[a,b]$ can be well approximated by elements from $C_0[a,b]$.

Even after all the reading, I am still unsure about how to begin and finish this proof. By the way, I am using the textbook Functional Analysis An Elementary Introduction by Haase.
Any help will be much appreciated, thanks in advance for your patience.

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1 Answer 1

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  1. Consider a sequence $(e_n)$ of functions in $C_0[a,b]$ such that $0\leq e_n\leq 1$ and $e_n(x) = 1$ if $a+1/n \leq x\leq b-1/n$. Then, for any $f\in C[a,b]$, check that $fe_n \to f$ in the $L^2$ norm.

  2. Prove that $C_0[a,b]$ is actually closed in the sup norm, and so it cannot be dense in $C[a,b]$.

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  • $\begingroup$ Thank you for the reply! In order to check that $fe_n \to f$ in the $L^2$ norm, then by definition, does this mean that $\forall \epsilon >0, \exists N >0 \quad \forall n \geq N \quad \left[ \int_a^b |fe_n - f|^2 dx \right]^{\frac{1}{2}} < \epsilon^2$? $\endgroup$
    – Jamil_V
    Feb 19, 2015 at 15:53

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