Tringular inequality proof for norm

Question

I can prove $L^2$ norm triangular inequality using the Cauchy-Schwarz inequality.

But how to prove the result for given $L^2$ space norm.

$||f||=\left(\int_0^1k(s)|f(s)|^2ds \right )^{1/2}$

where $\inf k(s)>0$ and $k$ is a continuous function.

How to show triangular inequality. I was not able to show using just the Cauchy-Schwarz inequality

Any Help will be appreciated

Thanks a lot

Answer 1

If you know the result for the usual $L^2((0,1))$ space than you can use this to prove your claim. I will assume everything is real valued and that your function $k$ is bounded. Furthermore I will follow your notation and write $||\cdot||$ for the weighted $L^2$ norm and $||\cdot||_{L^2((0,1))}$ for the usual $L^2$ norm. \begin{align*} ||f+g||&=\bigg(\int_0^1 k(s) |f(s)+g(s)|^2ds\bigg)^{1/2}\\ &=\bigg(\int_0^1|k(s)^{1/2}f(s)+k(s)^{1/2}g(s)|^2ds\bigg)^{1/2}\\ &=||k^{1/2}f+k^{1/2}g||_{L^2((0,1))}\\ &\leq ||k^{1/2}f ||_{L^2((0,1))}+||k^{1/2}g||_{L^2((0,1))}\\ &=||f||+||g|| \end{align*}