# $\delta : C([0,1])\to\mathbb{R}$, $f\mapsto f(0)$, with sup-norm is bounded but unbounded with one-norm.

The following is a problem out of Hunter's Applied Analysis that I am working on. I am looking for proof verification for the first part, and any constructive feedback/hints/tips for the second part of the question.

Let $\delta : C([0,1])\to\mathbb{R}$ be the linear functional that evaluates a function at the origin $\delta(f)=f(0)$. If $C([0,1])$ is equipped with the sup-norm, $$\|\ f \|_{\infty} = \sup_{0\leq x\leq 1} |f(x)|,$$ show that $\delta$ is bounded and compute its norm. If $C([0,1])$ is equipped with the one-norm, $$\|\ f \|_1 = \int_0^1 |f(x)|dx,$$ show that $\delta$ is unbounded.

First recall that for normed spaces $(X, \|\ \cdot \|_X)$ and $(Y, \|\ \cdot \|_Y)$ the norm of a linear operator, $T:X\to Y$ is defined as: $$\|\ T \|_{\text{op}} = \sup_{\|\ x \|_X = 1} \|\ Tx \|_Y$$ So in our example we're tasked with finding $$\|\ \delta \|\ = \sup_{\|\ f \|\ = 1} \|\ \delta(f) \|\ = \sup_{\|\ f \|\ = 1} |f(0)|$$ Then, note that $$\|\ \delta \|\ =\sup_{\|\ f \|\ = 1} |f(0)| \leq \sup_{\|\ f \|\ = 1 } \left( \sup_{0\leq x\leq 1}|f(x)| \right) = \sup_{\|\ f \|\ = 1 } \|f\|_{\infty} = 1$$ this establishes that $\delta$ is bounded. As far as the second bit of the question, consider the norm as: $$\|\ \delta \|\ = \sup_{\|\ f \|_1 \neq 0} \dfrac{\|\ \delta(f) \|\ }{\|\ f \|_1 } = \sup_{\|\ f \|_1 \neq 0} \dfrac{|f(0)|}{\|\ f \|_1 }$$ So I want to show the existence of an $f\in C([0,1])$ such that $f(0)= c > 0$ and $\int_0^1 |f(x)| = 0$. My first thought was $\cos(2\pi x)$ but the absolute value messes that example up. Anyway, needless to say, I'm stuck here. Any tips or hints?

You're not gonna find such a continuous function. as there clearly needs to be an interval of positive measure on which $f>0$ which would cause the integral to be nonzero.
Rather, take any sequence $f_n$ of area preserving functions such that $f_n(0)\rightarrow\infty$. As an example let $f_n(x)=2n(1-nx)$ on $x\in [0,1/n]$ and 0 elsewhere. Then the integral of $|f_n|$ is $1$ and $f_n(0)\rightarrow\infty$ as required.
You cannot find such an $f$. Rather, you are trying to find a sequence of $f_n$ with $\| f_n \|_{1} = 1$ and $\delta(f_n) \rightarrow \infty$. So think about this: how you make $f_n$ "big at $0$" but with a bounded integral?