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I have three linear mappings:

\begin{equation}t_0(f)=f(t_0)\end{equation}

\begin{equation}I(f)=\int_{0}^{1}f(t)f_0(t)dt\end{equation}

\begin{equation}T(f)=f(t)f_0(t)\end{equation}

and I want to determine whether or not they are continuous on $(C[0,1],\|\centerdot\|_1)$.

I have been trying to prove continuity by showing boundedness, e.g. $|T(f)|\leq M\|f\|_1$, with no success. I also have tried to construct a sequence $f_n$ satisfying $\|f_n\|_1=1$ and $|f_n(t)|\rightarrow\infty$, or find a sequence $f_n$ such that $\int_{0}^{1}f_n(t)dt\rightarrow0$ but $|T(f)|\nrightarrow0$. I have a hard time with counter examples.

I would greatly appreciate any hint or push in the right direction.

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Is $f_0(t)$ some fixed parameter in the definitions of $I(f)$ and $T(f)$? What does it mean? –  Patrick Apr 10 '12 at 4:04
    
I understood it to be just another function in C[0,1]. It says for any given $f_0\epsilon C[0,1]$. –  Ashley Apr 10 '12 at 4:24
    
Is the last problem a mapping into $(C[0,1],\|\centerdot\|_1)$? –  copper.hat Apr 10 '12 at 5:12
    
I assume so. The only information given is that the first two are linear functional and the third is just a linear operator. –  Ashley Apr 10 '12 at 13:45

1 Answer 1

up vote 2 down vote accepted

$||f||_1$ is the area under the curve $t \mapsto |f(t)|$. One way of thinking about the first problem is to look for a sequence of simple shapes that have constant area, but the height goes to $\infty$. Rectangles are an obvious choice, except they are not continuous, but this can be fixed easily in many ways.

For the second, consider using Hölder's inequality.

For the third (assuming this is a mapping into $(C[0,1],\|\centerdot\|_1)$), notice that $||Tf||_1 = I(Tf)$, where $I$ is from the second problem.

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I did try the second one with the Cauchy Schwartz inequality but I didn't get it to work. It should work? Thank you for your response! –  Ashley Apr 10 '12 at 15:45
1  
You need the generalization of the Cauchy Schwartz inequality which is $||f g ||_1 \leq ||f||_p ||g||_q$, where $1\leq p,q \leq \infty$ and $\frac{1}{p}+\frac{1}{q} = 1$. In this case, choose $p=1, q=\infty$. –  copper.hat Apr 10 '12 at 16:20

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