How do I show that $\sup_{0\leq x\leq1}|g(x)| \geq \int_0^1|g(x)|dx$

Question

I want to show that

$\sup_{0\leq x\leq1}|g(x)| \geq \int_0^1|g(x)|dx$

for real-valued $g$ that is continuous for $0\leq x\leq1$ .

Is it enough to say that

$|g(x)| \leq \sup_{0\leq x\leq1}|g(x)|$

and then that

$\int_0^1|g(x)|dx \leq \int_0^1\sup_{0\leq x\leq1}|g(x)|dx = (\sup_{0\leq x\leq1}|g(x)|)(\int_0^1dx)=\sup_{0\leq x\leq1}|g(x)|$

or is there some obvious flaw that I have missed?

Answer 1

We want to show

$\sup_{0\leq x\leq1}|g(x)| \geq \int_0^1|g(x)|dx$

for real-valued $g$ that is continuous for $0\leq x\leq1$ .

It is clear that

$0 \leq |g(x)| \leq \sup_{0\leq x\leq1}|g(x)|$

and then we see that

$\int_0^1|g(x)|dx \leq \int_0^1\sup_{0\leq x\leq1}|g(x)|dx = (\sup_{0\leq x\leq1}|g(x)|)(\int_0^1dx)=\sup_{0\leq x\leq1}|g(x)|$ .

Thus, the desired result is shown to be true.