Is this an inner product on $L^1$?

I know that $\int f(x) \overline{g(x)} dx$ is an inner product on $L^2$. But is it one on $L^1$? I think it isn't, but I am have had difficulty figuring out which defining property is violated.

Thanks in advance for any pointers!

Solution 1:

More explicitely: Take $L^1([0,1])$ and consider $f(x) = 1/\sqrt{x}$. Then $f\in L^1$ but $\int_0^1 f(x)f(x)dx = \int_0^1 f(x)^2 dx = \int_0^1\tfrac1x dx$ is infinite.

Solution 2:

Take $f$ a function in $L^1$ but not in $L^2$. We had examples recently-they blow up too much near a point. Then $\int f(x) \overline{f(x)} dx$ is not defined.

Solution 3:

Suppose that $g$ is a measurable function on $[0,1]$ such that $\int |fg|<\infty$ for all $f\in L^1[0,1]$. Then $g$ is essentially bounded.

If $g$ is not essentially bounded and $E_n=\{x\in[0,1]:n\leq |g(x)|\lt n+1\}$ for each positive integer $n$, then there is a subsequence $E_{n_1},E_{n_2},\ldots$ with $m(E_{n_k})\gt 0$ for all $k$. Let $\displaystyle{f=\sum_{k=1}^\infty}\frac{1}{k^2m(E_{n_k})}\chi_{E_{n_k}}$, so $f$ is in $L^1$, and $\displaystyle{\int |gf| \geq \sum_{k=1}^\infty\frac{n_k}{k^2}\geq\sum_{k=1}^\infty}\frac{1}{k}=\infty$.