1.Convergence in $L^1$

Question

Let f be measurable function such that $||f||_\infty=\infty.$ Show that there exists {${g_ n}$} $\subset L^1$ such that $||fg_n||_1\to\infty$. Anyhelp would be appreciated..

Answer 1

If $L^{1}=L^1(\Omega)$ where $\Omega$ is $\sigma$-finite, then $(L^{1})^*=L^\infty$, and if your claim didn't hold, there would be an $M>0$ such that for all $g\in L^1$, $\|fg\|_{L^1}<M$, and so $g\mapsto\int fg$ would be in $(L^1)^*$, which is impossible.

So under the assumption that your measure space is $\sigma$-finite, we can actually find a bounded sequence of $(g_n)$ in $L^1$ that work.

I do not think that this holds if $\Omega$ is not $\sigma$-finite. Consider the measure $\mu(A)=0$ if $A=\emptyset$ and $\infty$ otherwise on $\mathbb{R}$. Let $f(x)=x$, so $\|f\|_\infty=\infty$, but the only integrable function is the one which is identically $0$.

Answer 2

If you are working in $\mathbb{R}$:

Without loss of generality assume that $f,g \geq 0$. For each $n \in \mathbb{N}$ let $A_n \subseteq \mathbb{R}$ such that $0<\mu(A_n)<\infty$ and $f>n$ on $A_n$, where I use $\mu$ to denote Lebesgue measure on $\mathbb{R}$.

Let $g_n:= 1_{A_n}/\mu(A_n) \in L^1$. Then

$$\|fg_n\|_1 = \frac{1}{\mu(A_n)}\int_{A_n} f d\mu \geq n \rightarrow \infty \quad \textrm{as} \quad n \rightarrow \infty.$$

EDIT If $\mu(A) = \infty$, just note that

$$ A = \bigcup_{i \in \mathbb{Z}} (A \cap [i,i+1)) .$$

If all the sets in the union were null, then $A$ would also be null (as a union of null sets). Therefore at least one of these sets is non-null, say the one corresponding to $i = i_0$. But then

$$ 0 < \mu (A \cap [i_0,i_0 +1)) \leq \mu([i_0,i_0+1)) = 1 < \infty.$$

So we can just use the set $A \cap [i_0,i_0 +1)$ in place of $A$.