# An estimate involving Polar Coordinates

I encountered the following computation in a paper:

The highlighted $$t$$ is mysterious to me, how did it end up there? For some guidance: the first line is fundamental theorem of calculus and change of variables. The last equality is change of variables and fubini. Thanks!

• An upper bound $(\int_0^tf(r)dr)^2\le t\int_0^t[f(r)]^2dr$, where in this case $f(r):=\nabla u(r\xi)\cdot\xi$, might make sense iif we know more about $u$.
– J.G.
Commented Jan 11, 2021 at 14:38
• I think this is the direction to head. Assuming this is so, can you discuss this inequality in the answer section? Commented Jan 11, 2021 at 14:57
• Is "H" a Hermite polynomial or a Struve function? Commented Jan 11, 2021 at 15:11
• @K.defaoite no, its just shorthand for a type of measure on the surface of the sphere. "n-1 dim Hausdorff measure" to be more precise. Commented Jan 11, 2021 at 15:22

Jensen’s inequality should do the trick. Replacing $$dr$$ with the probability measure $$dr/t$$ on the interval $$[0,t]$$ and using convexity of $$x \mapsto x^2$$ furnishes the inequality $$\left( \int_{[0, t]} f(r) \frac{dr}{t} \right)^2 \leq \int_{[0, t]} f(r)^2 \frac{dr}{t}.$$ Pulling out the factors of $$t$$ and rearranging gives the desired inequality, $$\left( \int_{[0, t]} f(r) dr \right)^2 \leq t \int_{[0, t]} f(r)^2 dr$$ where $$f(r) = \nabla u (r \xi) \cdot \xi$$.