If the integral of $x(t) \cdot y(t)$ on the interval $(a,b)$ converges for all $x \in L^1(a,b)$ then $y$ must be in $L^{\infty}(a,b)$, that is $\text{supess}|y|< inf$?
Idea is to use Banach Steinhaus for the sequence of functionals :
$$A_n : L^1(a,b) \to \mathbb{C}; A_n(x)= \int_a^b x(t) y_n(t) \, dt$$ where $$y_n(t) := \begin{cases} y(t), & |y(t)| < n, \\ 0, & \text{otherwise} \end{cases}$$
But I can't write the proof step by step.
Can anyone help?
Define for each $n \in \mathbb{N}$
$$I_n(x) := \int_a^b x(t) y(t) 1_{\{|y| \leq n\}}(t) \, dt.$$
Then $I_n:L^1 \to \mathbb{C}$ is a linear operator and, since
$$|I_n(x)| \leq n \|x\|_{L^1},$$
the operator is also continuous. Moreover, since by assumption $x \cdot y \in L^1$ for any $x \in L^1$, it follows directly from the dominated convergence theorem that
$$I_n(x) \to I(x) := \int_a^b x(t) y(t) \, dt$$
for each $x \in L^1$. Applying Banach-Steinhaus yields that $I$ is also bounded, i.e. $\|I\|<\infty$. Now consider
$$x(t) := 1_{\{|y| \geq R}(t) \cdot \text{sgn}(y(t)).$$
for fixed $R>\|I\|$. Then
$$\int x(t) y(t) \, dt \geq R \int 1_{\{|y| \geq R}(t) \, dt = R \|x\|_{L^1}.$$
On the other hand,
$$\left| \int x(t) y(t) \, dt \right| \leq \|I\| \, \|x\|_{L^1}.$$
Hence, $$R \|x\|_{L^1} \leq \|I\| \, \|x\|_{L^1}.$$
Since we have chosen $R>\|I\|$, this inequality can only hold true if $\|x\|_{L^1} = 0$. This, in turn, implies that the set $\{t; |y(t)| \geq R\}$ is a null set; hence $\|y\|_{L^{\infty}} \leq R$.