Calculate the norm $A(f)$ in $C_{\mathbb{R}}[0,1] \rightarrow \mathbb{R} $

Question

$C_{\mathbb{R}}[0,1]$ is equipped with sup norm. Define $A$: $C_{\mathbb{R}}[0,1] \rightarrow \mathbb{R} $ as follow: $A(f)=f(0)+f(1)$. is this true that:

• $A$ is continuous since $|A(f)|=|f(0)+f(1)|\leq 2\cdot\sup\{{|f(x)|,x\in[0,1]\}}$

• And how to calculate the norm $||A(f)||$ ? Is it $||A(f)||=||f(0)+f(1)||=\sup\{f(0)+f(1)\}=f(0)+f(1)$

Answer 1

Note that $|f(t)| \le \|f\|$ for any $t$.

Hence $|Af| \le 2 \|f\|$.

Now choose a function $f$ such that $|f(0) + f(1)| = 2\|f\|$. There are lots of them, try with the simplest class of functions first (constants).

Answer 2

Hint: By definition, $\|A\|$ (or as you've written it, $\|A(f)\|$) is the lowest number $C$ such that for every $f$, $\|Af\| \leq C \|f\|$, where $\|f\| = \sup_{[0,1]}|f(x)|$.

You've already showed that $\|Af\| \leq 2\|f\|$, which is to say that $C = 2$ works, which is to say that $\|A\|$ is at most $2$. Now, you need to answer the question: why is there no lower number (that is, no value of $C$ lower than $2$) such that $\|Af\| \leq C \|f\|$ for every $f$?

Answer 3

1. For any normed linear space $X$ over $\mathbb R$ or over $\mathbb C,$ a linear function $A:X\to \mathbb R$ or $A:X\to \mathbb C$ is continuous iff it is Lipschitz-continuous iff $\exists M\in \mathbb R^+\;\forall f\in X\;(|A(f)|\leq M\|f\|).$

2. Def'n: $\|A\|=\sup \{|A(f)/\|f\|: f\ne 0\}$ for continuous $A.$

3. For continuous $A \ne 0:$ Let $$m_A=\{M>0: \forall f\in X\;(|A(f)|\leq M\|f\|\}.$$ Let $N_A=\inf m_A.$ Then $N_A>0$ (else $A=0$). By the def'n of $\inf$, whenever $e\in (0, N_A)$ there exists $g\in X$ with $|A(g)|>(N_A-e)\|g\|.$ For such $g$ we must have $g\ne 0.$ Therefore $$N_A\leq \sup \{|A(g)|/\|g\|:g\ne 0\}=\|A\|.$$ But if there was some $f\ne 0$ with $|A(f)|/\|f\|>N_A$ then by def'n of $N_A$ we would have $N_A\geq |A(f)|/\|f\|>N_A$ , which is absurd. So we have $$\|A\|=\sup \{|A(f)/\|f\|: f\ne 0\}\leq N_A.$$ So $N_A\leq \|A\|\leq N_A,$ so $N_A=\|A\|.$

4. In your Q you have $|A(f)|\leq 2\|f\|$ for all $f$ so $A$ is continuous. And $2\in m_A$ so $\|A\|=N_A=\inf m_A\leq 2.$

   To show that $\|A\|=2$ it suffices now to show that $\|A\|\geq 2.$ To do this, find $f\in C[0,1]$ with $f(0)=f(1)=\|f\|=1.$ Then $$\|A\|=\sup \{|A(g)|/\|g\|:g\ne 0\}\geq |A(f)\|/\|f\|=2.$$ For example if $f(t)=1$ for all $t\in [0,1].$

Remark. Even when $X$ is a separable Banach space whose dual space $X^*$ is separable, it is not always the case that, whenever $0\ne A\in X^*,$ that some $f\in X$ exists with $|A(f)|=\|A\|\cdot \|f\|\ne 0$.