$d(f,g)=\int_X |f-g|^p d\mu$ defines metric space when $0<p\leq 1$

Question

I have tried solving this problem.

Show that $$d(f,g)=\int_X |f-g|^p d\mu$$ defines metric space when $0<p\leq 1$.

The only thing I need to show is triangular inequality.

I have tried to use Holder inequality but it does not work since $p\leq 1$. So I think, I need to prove that

$$|f-h|^p\leq |f-g|^p+|g-h|^p$$

directly. Buy I don't know how... seems very easy... but ....

Any hint or answer would be helpful! Thanks in advance

Answer 1

Using the triangle inequality, you can expand as:

$$ |f-h|^p = |f-g+g-h|^p \leq (|f-g|+|g-h|)^p $$

And then you can use the result given by this question to get your result.