# Norm of a linear functional

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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Aaron: My mathematical background is not enough to fully understand your second paragraph. However, your concrete example in the third paragraph is just great. I'm amazed by this way of approximating the sign function; very, very useful. Thanks! –  ragrigg Oct 15 '12 at 19:07
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. –  Aaron Oct 15 '12 at 22:08
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.