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Thread: not continuous linear functional

  1. #1
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    not continuous linear functional

    Hi again! We are in a $\displaystyle L^p$-space with $\displaystyle 0<p<1$ ($\displaystyle L^p=\{f:[0,1] \rightarrow \mathbb{K}: f $L-measurable, $\displaystyle \int_0^1 |f(t)|^p dt < \infty\}$ with $\displaystyle d(f,g):= \int_0^1 |f(t)-g(t)|^p dt$.
    We consider a linear functional $\displaystyle \phi \neq 0$. Let $\displaystyle f \in L^p: \int |f|^p =1, \phi(f)=\alpha > 0$. Let $\displaystyle F(t):=\int_0^t |f(s)|^p ds$. Then F is continuous, therefore attains all values $\displaystyle \frac{k}{n}$, say at $\displaystyle s_k, k=0,...,n.$ Define $\displaystyle g_r:=f \cdot c_{[s_{r-1},s_r]}$ (c meaning the char. function). There is a $\displaystyle r(n)$, so that $\displaystyle |\phi(g_{r(n)})| \geq \frac{\alpha}{n}.$ Using $\displaystyle f_n:=n \cdot g_r(n)$ one can now show that $\displaystyle \phi$ is not continuous. But how?
    I do not see why that follows. Does anybody know and can help me?
    Last edited by Recursion; Jan 13th 2009 at 12:46 PM.
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