I encountered the following computation in a paper:

enter image description here

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!

  • $\begingroup$ 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 $u$. $\endgroup$
    – J.G.
    Commented Jan 11, 2021 at 14:38
  • $\begingroup$ I think this is the direction to head. Assuming this is so, can you discuss this inequality in the answer section? $\endgroup$
    – yoshi
    Commented Jan 11, 2021 at 14:57
  • $\begingroup$ Is "H" a Hermite polynomial or a Struve function? $\endgroup$
    – K.defaoite
    Commented Jan 11, 2021 at 15:11
  • $\begingroup$ @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. $\endgroup$
    – yoshi
    Commented Jan 11, 2021 at 15:22

1 Answer 1


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 $f(r) = \nabla u (r \xi) \cdot \xi$.

  • $\begingroup$ haha, still don't see it. could you provide a little more detail? thanks! $\endgroup$
    – yoshi
    Commented Jan 11, 2021 at 18:41
  • 1
    $\begingroup$ @yoshi edited with more detail $\endgroup$
    – user574443
    Commented Jan 11, 2021 at 19:50

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