Quantcast
Channel: Active questions tagged real-analysis - Mathematics Stack Exchange
Viewing all articles
Browse latest Browse all 8465

Showing a lower bound for the square root of a function

$
0
0

Suppose that we have a function $f$ that is of exponential type, and that the function $g(x) = |f(x-i)|^2$ is such that it satisfies $$\left(\frac{1}{|B|} \int_B g(x)\right)\left(\frac{1}{|B|} \int_B g(x)^{-1}\right) \leq C < \infty,$$ for all balls $B$. Can I conclude that there is some lower bound for $g$? Apparently there is a way to bound it in the sense of $$\sqrt{g(x)} = |f(x-i)| \geq \frac{C}{1 + |x|^3},$$ for all $x \in \mathbb{R}$.

How is this possible? I assume we would have to proceed by contradiction somehow and essentially arrive to the impossible scenario where the inequality above is false for some appropriately chosen $B$. But I am not sure how. Any hint is welcomed!

Edit: This claim is from a paper I read, I am more interested in the claim itself and not the result in the paper you can find it on page 8 here.


Viewing all articles
Browse latest Browse all 8465

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>