Inequality for product of integrals

Question

Consider two real-valued functions $g: \mathbb{R}\rightarrow \mathbb{R}$ and $f:\mathbb{R}\rightarrow \mathbb{R}$. Is this relation true:

$$ \int_{\mathbb{R}}g(x)f(x)dx\leq \sup_x f(x)\int_{\mathbb{R}} g(x)dx $$

? Why?

Answer 1

If you restate that inequality with absolute values then it is true. Consider the Riemann integral in terms of the limit of a series. Let $M=sup\{f(x):x \in \mathbb{R}\}$ and $a<b$ be arbitrary points on the real line. Define $x_i=a+i/n$. Then:

$$\int_{a}^{b}|g(x)|f(x)dx=\lim_{n \to \infty} \sum_{i=1}^nf(x_i^*)|g(x_i^*)|[x_i-x_{i-1}]$$

$$\leq\lim_{n \to \infty} M\sum_{i=1}^n|g(x_i^*)|[x_i-x_{i-1}]$$

$$=M\int_{a}^{b}|g(x)|dx$$

As this is true for an arbitrary interval $[a,b]$ it is true for the whole Real line.

Answer 2

Let $f = \mathbb{1}_{[1,2]}$ and $g = - \mathbb{1}_{[0,1]}$ .

Then $f(x)g(x) = 0$ for all $x \in \mathbb R$, so the LHS is $0$, whereas the RHS is $1 \cdot (-1) = -1$.

However, if $g(x) \geq 0$ for all $x \in \mathbb R$, and if $f$ is bounded from above, then the inequality holds.