prove that a measurable function is in $L^{\infty}([0,1])$

Question

I am trying to solve this problem but I can not find a good idea:

Let $\phi$ be a measurable function on $[0,1]$ and assume that the linear transformation

A $\colon f \to f \cdot \phi$

maps $L^{2}([0,1])$ into itself .Prove that $\phi \in L^{\infty}([0,1])$.

I tried to prove it by contradiction by choosing properly a function f so that $f\phi$ is no longer in $L^2([0,1])$ but I was no able to find any of it. Any hint for this problem would be appreciated.

Answer 1

The condition is that $$\int_{[0,1]}|\varphi|^2\,|f|^2<\infty$$ for all $f\in L^2$. So the operator $A_0:f\longmapsto |\varphi|\,f$ also maps $L^2$ into $L^2$.

We also have $$ \int_{[0,1]}(1+|\varphi|)^2\,|f|^2\leq 2\int_{[0,1]}(1+|\varphi|^2)\,|f|^2\leq\int_{[0,1]}|f|^2+\int_{[0,1]}|\varphi|^2\,|f|^2<\infty, $$ so the operator $T:f\longmapsto (1+|\varphi|)f$ also maps $L^2$ into $L^2$. If we define $S:f\longmapsto \tfrac1{1+|\varphi|}\,f$, then $$ \|Sf\|_2^2=\int_{[0,1]}|Sf|^2\leq\int_{[0,1]}|f|^2=\|f\|_2^2, $$ so $S$ is bounded. As $TS=ST=I$, we have the $S$ is bounded and bijective. Then $T$ is bounded by the Bounded Inverse Theorem (that is, because $S$ is bounded it is open by the Open Mapping Theorem, and that makes $T$ continuous). Knowing that $T$ is bounded, we get that $A_0=T-I$ is bounded. Since $$ \|Af\|_2^2=\int_{[0,1]}|\varphi|^2\,|f|^2=\|(A_0-I)f\|^2_2 $$ for all $f$, we get that $\|A\|=\|A_0-I\|$.

Now let $L=\{|\varphi|>\|A\|+1\}$. Then, if $m(L)>0$, \begin{align} \|A\|^2\,m(L)&=\int_{[0,1]}\|A\|^2\,1_L^2\leq\int_{[0,1]}(|\varphi|-1)^2\,1_L^2<\int_{[0,1]}|\varphi|^2\,1_L^2\\[0.3cm]&=\|A\,1_L\|_2^2\leq\|A\|^2\,\|1_L\|_2^2=\|A\|^2\,m(L). \end{align} a contradiction. So $m(L)=0$, making $\varphi$ essentially bounded. You can actually do the above with $\delta>0$ instead of $1$, to end up showing that $\|A\|=\|\varphi\|_\infty$.

Answer 2

Hint 1: Replacing $f(x)$ by $f(x){\bar{\phi}(x) \over |\phi(x)|}$ you can assume $\phi(x)$ is nonnegative.

Hint 2: Let $\phi_n(x) = \phi(x) $ when $n < \phi(x) \leq n+1$ and $\phi_n(x) = 0$ otherwise. Apply the uniform boundedness principle to $A_n f(x) = \phi_n(x) f(x)$.

Answer 3

Here is a simple way to prove the result by contradiction.

Let $\phi$ be a measurable function on $[0,1]$. Suppose that $\phi \notin L^{\infty}([0,1])$. We are going to define a function $f \in L^2([0,1])$, such that $f\phi \notin L^2([0,1])$.

For each $n \in \mathbb{N}$, let $E_n= \{ x \in [0,1] : n \leqslant |\phi(n)| < n+1 \}$. Note that if $n\neq m$ then $E_n$ and $E_m$ are disjoint.

Now, let $S = \{n \in \mathbb{N} : n>0 \textrm{ and }\mu(E_n)>0 \}$. Since $\phi \notin L^{\infty}([0,1])$, $S$ is an infinite set.

Define $f$ by $$ f = \sum_{n \in S} \frac{1}{n\sqrt{ \mu(E_n) }} \chi_{E_n} $$

where $\chi_{E_n}$ is the characteristic function of $E_n$.

It is easy to see that

$$ \int_{[0,1]} |f|^2 = \sum_{n \in S} \frac{1}{n^2} <+\infty $$

So $f \in L^2([0,1])$.

However,

$$ \int_{[0,1]} |f\phi|^2 = \int_{[0,1]} |f|^2|\phi|^2 \geqslant \sum_{n \in S} \frac{1}{n^2 \mu(E_n) } n^2 \mu(E_n)=\sum_{n \in S} 1 = +\infty$$

So $f\phi \notin L^2([0,1])$.