This may be a simple enough question but is there an equivalent of the triangle inequality when there is a power of $1/2$ and $x$ and $y$ are complex-valued functions. That is \begin{equation} |x + y|^{1/2} \leq C\left( |x|^{1/2} + |y|^{1/2}\right) \end{equation} where $C$ is a positive constant. I saw a result similar to this at Triangle inequality for higher powers but the result there only works for powers greater of equal to $1$.
Asked
Active
Viewed 93 times
0
-
5$|x+y|^{1/2}\leq (|x|+|y|)^{1/2}\leq |x|^{1/2}+|y|^{1/2}$. – geetha290krm Jun 02 '23 at 09:32
-
2https://math.stackexchange.com/q/408177/42969, https://math.stackexchange.com/q/583415/42969 – Martin R Jun 02 '23 at 09:32
-
Fantastic, thanks! – Jason Curran Jun 02 '23 at 09:38