Consider a functional $C([0,1])\ni f \mapsto f(1) \in \mathbb{R}$. Let the norm on $C([0,1])$ be:
$$\|f\| = \sup\limits_{x\in[0,1]}|f(x)| + \int\limits_0^1 |f(x)|dx. $$
Then $|f(1)|\leq \sup\limits_{x\in[0,1]}|f(x)|\leq \sup\limits_{x\in[0,1]}|f(x)| + \int\limits_0^1 |f(x)|dx = \|f\|$, which proves that the functional is continuous. I would like to show that the norm of the functional is $1$. A function in $C([0,1])$ which lies in the unit sphere with respect to the given norm and its value at $1$ is equal to $1$ does not exist(at least I think so). So I have constructed a sequence of functions $\{f_n\}_{n=1}^{\infty}$ such that:
- $f_n(1) = 1-\frac{1}{n}$.
- $\forall n$ exists $\delta$ s.t. $f_n(x) = 0$, for $x<1-\delta$.
- $f_n(x)<f_n(1)$ for $x\in [0,1)$.
- $\int\limits_0^1 |f_n(x)|dx = \frac{1}{n}$
Then $\|f_n\| = 1$ for all $n$ and $|f_n(1)| = 1 - \frac{1}{n} \rightarrow 1$ as $n \rightarrow \infty$.
Would such an approximation be sufficient?