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I'm doing a practice exam for Real Analysis and am wondering about this specific question:

Let $C[-1,1]$ be the set of all real-valued continuous functions on $[-1,1]$. For $f,g\in C[-1,1]$ and $\lambda\in\mathbb{R}$, define $$(f+g)(t)=f(t)+g(t),\,(\lambda f)(t) = \lambda f(t),\,t\in[-1,1].$$ Then $C[-1,1]$ is a vector space over $\R$ with the given operations. For $f\in C[-1,1]$, define $$\|f\|=\max\{|f(t)|\,|\, t\in[-1,1]\}.$$ (b) Define $\ell:C[-1,1]\to\R$ by $\ell(f)=\int_{-1}^0 f(t)\,dt-\int_0^1 f(t)\,dt$ for $f\in C[-1,1]$. Prove that $\ell$ is a bounded linear function and find $\|\ell\|$.

I think that $\|\ell\|=2$, but I'm not 100% sure and I want to see if I'm right. I got so far that $\|\ell\|\leq 2$, because if we let $\|f\|=1$ we find $$\begin{align*} |\ell(f)|&=\left|\int_{-1}^0 f(t)\,dt-\int_0^1 f(t)\,dt\right|\\ &\leq \left|\int_{-1}^0 f(t)\,dt \right|+\left|\int_0^1 f(t)\,dt\right|\\ &\leq \int_{-1}^0 |f(t)|\,dt+\int_0^1 |f(t)|\,dt\\ &\leq \int_{-1}^0 1\,dt+\int_0^1 1\,dt\\ &=2. \end{align*}$$

I'm thinking that to show $\|\ell\|\geq 2,$ I can use the sequence $\{f_n\}$ in $C[-1,1]$ defined by $$f_n(t)=\begin{cases} 1 & \text{if }-1\leq t\leq -1/n,\\ -nt & \text{if }-1/n<t<1/n,\\ -1 & \text{if }1/n\leq t\leq 1. \end{cases}$$ We find that for each $n$, $\|f_n\|=1$. Also, $|\ell(f_n)|=2-1/n$ so that $\lim_{n\to\infty}|\ell(f_n)|=2$. Does this mean $\|\ell\|\geq 2$?

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    $\begingroup$ Looks OK to me. This is an example of a linear functional on a normed vector space that does not attain its norm (i.e. there is no $f \in C[-1,1]$ with $\|f\| = 1$ and $\|\ell(f)\| = 2$), which perhaps "explains" why you are using a sequence of elements to show that $\|\ell\| \geq 2$. See e.g. the related problem math.stackexchange.com/questions/80773/… $\endgroup$ Commented Aug 18, 2023 at 0:20
  • $\begingroup$ @leslietownes yes that's exactly why I resorted to a sequence! I'm glad to know that my idea worked $\endgroup$ Commented Aug 19, 2023 at 5:47

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Your argument is fine. A slightly more abstract point of view is to notice that you have $$\tag1\ell(f)=\int_{-1}^1 fg,$$ where $$\tag2 g=1_{[-1,0]}-1_{(0,1]}. $$ For such functional as in $(1)$, it is not hard to show (using the usual ideas to show that $L^1$ is the dual of $C_0$ and $L^\infty$ is the dual of $L^1$) that $$\tag3 \|\ell\|=\|g\|_1^{\vphantom1}. $$ And in this case you have $\|g\|_1^{\vphantom1}=2$.

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  • $\begingroup$ When you say $L^1$ “is” the dual of $C_0$ do you mean that they’re isomorphic? $\endgroup$ Commented Aug 18, 2023 at 2:24
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    $\begingroup$ Isometrically isomorphic in a canonical way, as expressed by $(1)$. The same way that $L^\infty$ is the dual of $L^1$. $\endgroup$ Commented Aug 18, 2023 at 4:03

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