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Let $(X,\|\cdot \|)=(C[0,1],\|\cdot \|_{\infty})$ and $F_1 :X \rightarrow \mathbb R$ be defined by $$F_1 (f)=\int _{1/2}^{3/4} f(t) dt$$ $F_1$ is a continuous linear functional.

Lets consider the normed space $C^1[0,1]$ of all continuously differentiable real valued functions defined on $[0,1]$, with the supremum norm. Check that $F_2(f)=f'(1)+f(1)$ is a linear functional. Is it continuous ($\iff$ bounded)?


Let $a \in \mathbb R$ and $f,g \in C^1[0,1]$. So $$F_2(af+g) = (af+g)'(1) + (af+g)(1) = (af)'(1)+g'(1) + (af)(1) + g(1) $$ $$= af'(1)+g'(1) + af(1) + g(1) $$ $$= a(f'(1) +f(1)) + g'(1) + g(1) = aF_2(f) + F_2(g)$$ so it is a linear functional.

Is this correct? Are $f,g$ meant to be in $C'[0,1]$?

I don't know what to do in the continuous part of the question.

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It's unclear if you have any question about $F_1$; your questions seem to be about $F_2$ only.

Your proof of linearity looks fine.

It's easy to see that, given any natural number $n$, you can construct $f$ in $C^1$, as follows: $f(0)=0$ for $x\in[0,1-1/n]$, $f(x)$ rises smoothly from $0$ to ${1\over 2}$ for $x\in [1-1/n, 1-1/(2n)]$, and $f(x)$ rises linearly from ${1\over 2}$ to $1$ for $x\in [1-1/(2n),1]$. For such an $f$, we have $f'(x)=n$ for $x\in [1-1/(2n),1]$, and we also have $||f||_\infty=1$ but $F_2(f)=n+1$, and so it seems difficult for $F_2$ to be continuous.

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  • $\begingroup$ when you say it is difficult for it to be continuous, does that mean it is not continuous or it can be continuous...? $\endgroup$ – snowman Mar 20 '16 at 23:20
  • $\begingroup$ Apologies for being unclear. I assume you know that a linear functional on a normed linear space is continuous iff it is bounded. (Rudin Real and Complex Analysis 5.4, second edition.) If we can choose $f$ with $||f||_\infty=1$ but $F_2(f)$ as big as we want, then $F_2$ isn't bounded, and therefore isn't continuous. $\endgroup$ – ForgotALot Mar 20 '16 at 23:29

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