Question

h(t) is once differentiable function on [0,1] and h(0)=h(1)=0.
I'll write question in comment.

Answer 1

Why
\[\int\limits_{0}^{1}h'(t)dt=0\]
but
\[\int\limits_{0}^{1}h'(t)^{2}dt \ge0\]
?

Answer 2

Well the first one is easy right?
\[\int\limits_0^1 \frac{d h(t)}{dt}dt = h(1)-h(0); \text{but } h(1) = 0 =h(0); \text{so} h(1)-h(0)=0\]

Answer 3

ok

Answer 4

Well we know that:
\[0=\left| \int\limits_0^1 h(t)dt\right| \le \left| \int\limits_0^1 (h(t))^2dt\right|\]
For any function h(t), so since int of h(t)^2 is positive definite you can drop the magnitude and that implies the integral is greater than or equal to 0.

Answer 5

I feel as though these are the "qualitative' arguments that I get from my professors when it comes to stuff like this.

Answer 6

thank you :)

Answer 7

I have one more question.
Why
\[3\int\limits_{0}^{1}h'(t)^{2}(1+\frac{ 1 }{ 3 }h'(t)^{3})dt \ge0\]
if
\[||h||_{1}\le3\]
where
\[||h||_{1}=\max|h(t)|+\max|h'(t)|, t \in[0,1]\]