# Showing a linear mapping is continuous (or not)

I have three linear mappings:

$$t_0(f)=f(t_0)$$

$$I(f)=\int_{0}^{1}f(t)f_0(t)dt$$

$$T(f)=f(t)f_0(t)$$

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

$||f||_1$ 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 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.
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