# Linear forms of $L^1$

I am an engineering student who has unwittingly taken a module in functional anylysis which, unfortunately, is ever so slightly over my head. I would greatly appreciate if you could either point me in the right direction for the questions below, or at least tell me where I might find some materials which would enlighten me in this area.

So, $$\varphi(f)=\int u\times f$$$$u\in L^\infty, f\in L^1$$ Show that $\varphi(f)$ is a continuous linear form on $L^1$.

Now, if $$A=\text{sup}{ \{ |\varphi(f)| \text{ for } f\in L^1 \text{ and } ||f||_1 \leq1 \}}$$ can $A=\infty$?

Many thanks.

-
There are some details missing for this problem, such as what space these functions $f$ and $u$ are $L^1$ and $L^{\infty}$ on exactly, and what the space of the codomains of these functions are. But I think this can get you started: A linear operator is bounded if and only if it is continuous. Thus, just show that $\varphi$ is a linear operator, and $\| \varphi (f) \|_1 \leq M \| f \|_1$ for some $M >0.$ –  Ragib Zaman Nov 13 '11 at 11:16
Holder inequality (I imagine you could do it more directly using the above comment) –  David Mitra Nov 13 '11 at 11:30

To show that $\varphi$ is continuous, you have to show that $A$ is finite. Let $M=||u||_\infty$. Then for $||f||_1\le 1$: $$|\varphi(f)|=\biggl|\int uf\,\biggr|\le\int|uf|\le \int M|f|=M\int |f|=M||f||_1\le M.$$
That $\varphi$ is linear follows from the linearity of integration.
Thank you David and Ragib, that was very useful. I should of course have mentioned that the space is $\mathbb{R}$. A later question is similar but for $L^2$: $\psi (f) = \varphi (w \times f), w,f \in L^2 (\mathbb{R})$. I assume I have to do the same process but for $L^2$ i.e show that $||\psi(f)||_2 \leq M||f||_2$? Thanks once again. –  NIM Nov 13 '11 at 17:33