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

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.
