Inequations about curvature tensor.

Question

How to get two inequations 1 and 2?

At the 1, I don't know how to use the Cauchy inequation .

At the 2, I absolutely start from where..

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Answer 1

Let's define $f = (BM^2 + |Rm|^2)$ and $h = |\nabla Rm|^2$ so that $S = fh$. Then the product rule gives $$\partial_t S = f \partial_t h + h \partial_t f$$ and $$\Delta S = f \Delta h + h \Delta f + 2 \langle \nabla f, \nabla h \rangle.$$

Using these and the given bounds for the time derivatives of $f,h$ we get

$$\begin{align} \partial_t S &\le h (\Delta f - 2 |\nabla Rm|^2 + C|Rm|^3) + f(\Delta h - 2|\nabla^2 Rm|^2 + C|Rm||\nabla Rm|^2) \\ &= \Delta S -2 \langle \nabla f, \nabla h \rangle + h (- 2 |\nabla Rm|^2 + C|Rm|^3) + f( - 2|\nabla^2 Rm|^2 + C|Rm||\nabla Rm|^2). \end{align}$$

Now use the product rule to get $\nabla f = \nabla|Rm|^2 = 2\langle Rm, \nabla Rm\rangle$ and similiarly $\nabla h = 2\langle\nabla Rm, \nabla^2 Rm\rangle$, so Cauchy-Schwarz gives $$-2 \langle \nabla f, \nabla h \rangle \le 8|Rm||\nabla Rm|^2 |\nabla^2 Rm|\le 8 M |\nabla Rm|^2 |\nabla^2 Rm|.$$

The first parenthetical is easy to estimate: $$ h (- 2 |\nabla Rm|^2 + C|Rm|^3) = -2 |\nabla Rm|^4 + C|\nabla Rm|^2 |Rm|^3 \le -2 |\nabla Rm|^4 + CM^3 |\nabla Rm|^2.$$

The second is not so bad either: if we estimate $BM^2 \le f \le (B+1)M^2$ we get $$f( - 2|\nabla^2 Rm|^2 + C|Rm||\nabla Rm|^2) \le -2BM^2|\nabla^2 Rm|^2 + C(B+1)M^3|\nabla Rm|^2.$$

Putting this all together and collecting like terms we get

$$ \partial_t S \le \Delta S -2BM^2 |\nabla^2 Rm|^2-2|\nabla Rm|^4 + 8 M |\nabla Rm|^2 |\nabla^2 Rm| + CM^3(B+2)|\nabla Rm|^2$$

This is not quite what we want - we have an 8 instead of a $C$, and $B+2$ instead of $B$. See if you can spot an error in my work or a way to use $B \ge C^2 / 4$ to fix it.