# An integral inequality question.

If we have two functions $f,g:[a,b]\to\mathbb{R}$ and we know they are bounded, so:

$\sup_{x\in[a,b]}|f(x)|=K$, and $\sup_{x\in[a,b]}|g(x)|=M$.

Where $K,M$, are positive finite constants, which of the following inequalities is the correct one?

$(1)$ $\int_a^b|f(x)||g(x)|dx\leq \sup_{x\in[a,b]}|f(x)|\sup_{x\in[a,b]}|g(x)|=KM$, or

(2) $\int_a^b|f(x)||g(x)|dx\leq \sup_{x\in[a,b]}|f(x)|\sup_{x\in[a,b]}|g(x)|\int_a^b1\,dx=KM(b-a).$

Thanks!

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Counterexample Let $f=g=1$ (constant) $a=0, b=2$ then $$\int_0^2|f(x)||g(x)|dx=\int_0^2dx=2$$ while $$\sup_{x\in [0, 1]}|f(x)|\sup_{x\in [0, 1]}|g(x)|=1.$$ Note that $$|f(x)|\leq \sup_{x\in [a, b]}|f(x)|\quad \text{and} \quad|g(x)|\leq \sup_{x\in [a, b]}|g(x)|$$ for all $x\in [a, b]$. It follows that $$\int_a^b|f(x)||g(x)|dx\leq \int_a^b\sup_{x\in [a, b]}|f(x)|\sup_{x\in [a, b]}|g(x)|dx =\int_a^b(MK)dx=MK(b-a)$$
and what if we just had $\int_a^b|f(x)|dx$, i.e. only one function in the integral? –  ellya Apr 7 at 9:33
@ella $\displaystyle\int_a^b|f(x)|dx\leq (b-a)\sup_{x\in [a, b]}|f(x)|$ –  impartialmale Apr 7 at 9:35