# Define $\ell:C[-1,1]\to\mathbb{R}$ such that $\ell(f)=\int_{-1}^0f-\int_0^1 f$ and let $\|f\|=\max_{t\in[-1,1]}|f(t)|$. Is $\|\ell\|=2$?

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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 $$f_n(t)=\begin{cases} 1 & \text{if }-1\leq t\leq -1/n,\\ -nt & \text{if }-1/n We find that for each $$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$$?

• 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/… Commented Aug 18, 2023 at 0:20
• @leslietownes yes that's exactly why I resorted to a sequence! I'm glad to know that my idea worked Commented Aug 19, 2023 at 5:47

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 $$\|g\|_1^{\vphantom1}=2$$.
• When you say $L^1$ “is” the dual of $C_0$ do you mean that they’re isomorphic? Commented Aug 18, 2023 at 2:24
• Isometrically isomorphic in a canonical way, as expressed by $(1)$. The same way that $L^\infty$ is the dual of $L^1$. Commented Aug 18, 2023 at 4:03