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Consider $C[0,1]$ (the space of continuous functions on $[0,1]$) with the max-norm (assume the underlying field is $\mathbb{R}$). For $g \in C[0,1]$, define $\Phi_g: C[0,1] \rightarrow \mathbb{R}$ by

\begin{equation*} \Phi_g(f) = \int_0^1 f(t)g(t) \space dt, \end{equation*}

where the integral is the ordinary Riemann integral.

I want to prove that $\Phi_g \in C[0,1]^*$ and $\| \Phi_g \|= \int_0^1 |g(t)| \space dt$. I've proved that

\begin{equation*} \| \Phi_g(f) \| \leq \| f \| \int_0^1 |g(t)| \space dt. \end{equation*}

Therefore, all I'm missing is an $f \in C[0,1]$ such that $\|f\| \leq 1$ and $\| \Phi_g(f) \| = \int_0^1 |g(t)| \space dt$. The constant functions $1$ or $-1$ work if $g$ is always positive or negative, respectively. Any idea about what function could do the job in any other case? My first idea was $f = |g|/g$, but this $f$ is not necessarily continuous.

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As Berci says, the idea is to approximate $|g|/g$ (where we define the function to be $0$ when $g(x)=0$) with continuous functions. Here is a way to explicitly construct a sequence of continuous functions which approximate $|g|/g$, followed by an explicit and concrete sequence.

The idea is to convolve $|g|/g=\operatorname{sign}(g)$ with a family of mollifiers, that is, continuous functions which have area $1$ and which approach the delta function. For example, let $f(x)=x+1$ if $-1\leq x\leq 0$, $f(x)=1-x$ if $0\leq x \leq 1$ and $0$ elsewhere. Define $f_k(x)=kf(kx)$. Then $\int f_k(x)dx=1$ and $f_k$ is supported on $[-1/k,1/k]$. If you define $h_k=|g|/g\star f_k$ (where $\star$ denotes convolution), then $h_k$ will be continuous and $h_k\to |g|/g$ This is a general technique that will give you approximations converging to a given function (and if you convolve with a smooth family, it will give smooth approximations). The disadvantage of this technique is that it's hard to say with certainty what the functions "look like".

A more concrete option is to take $h_n(x)=ng(x)$ if $|ng(x)|<1$ and $|g(x)|/g(x)$ otherwise. Then $h_n$ converges to $|g|/g$. Moreover, where $h_n\neq |g(x)|/g(x)$, we have $|g(x)|<1/n$, and so $|\int_0^1 g(x)(h_n(x)-|g(x)|/g(x))dx|<1/n$

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    A few stray comments. If you let $s_n(x)=nx$ when |x|<1/n and $x/|x|$ otherwise, we are using $s_n\circ g$. Second, if you are doing functional analysis, convolving with mollifiers will come up eventually, so it is worth looking into. Third, while Norbert's idea is overkill for this problem, I made essential use of the fact that we are on a finite interval. His technique extends to the whole line. It is worth internalizing his idea.2012-10-15
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Hint: Not necessarily one function is what you need. Of course, the target is the mentioned $|g|/g$, but you can approximate it by continuous functions.

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    I thought about that. Right now I don't see the desired sequence.2012-10-15