Prove that $\vert \int_{B_a}f(x)dx \vert \leq \frac{a^{m+1}}{m^2+m}\omega _m \sup_{B_a} |\nabla f|$

Question

Assume that $a\in \mathbb{R}^{+}, m\in\mathbb{N}^{+}, B_a \subset \mathbb{R}^m$. $f:\overline{B_a} \to \mathbb{R}$ is $C^1$, $f|_{\partial{B_a}}=0$. Prove that $$ \Big| \int_{B_a}f(x)dx \Big| \leq \frac{a^{m+1}}{m^2+m}\omega _m \sup_{B_a} |\nabla f|, $$ where $B_a=\{x\in \mathbb{R}^m:|x| \leq a \}$, $\omega_m$ denotes the area of m-dimensional unit ball, i.e., $\omega_m=\int_{\partial{B_1}} 1d\sigma$. In fact the area of m-dimensional ball with radius $r$ is $\int _{|x|=r} d\sigma$, which equals $\omega_m \cdot r^{m-1}$.
Hint from my professor: Apply the coarea formula(1), and the fact that if $\Omega \subset \mathbb{R}^m$ is an open region, $f \in C^1{(\Omega)}, [a,b] \subset \Omega$, then $$ f(b)-f(a)=\int_0 ^1 \nabla f(a+t(b-a))\cdot (b-a)dt.(2) $$ (1):Let $G \subset \mathbb{R}^m$ be bounded open region, $f \in C^1(G), \forall x \in G$ we have $\nabla f(x) \neq 0, \Omega = f^{-1}[a,b]$. Let $g\in C(\Omega)$, then $$ \int _\Omega g=\int _a ^ b dt \int _{f^{-1}(t)} \frac{g}{|\nabla f|} d\sigma. $$ My attempt: if we set $h:\mathbb{R}^m \to \mathbb{R}, x\mapsto |x|$, then $B_a=h^{-1}[0,a]$. Applying (1): $$ \int_{B_a}f(x)dx=\int_{h^{-1}[0,a]}f(x)dx=\int _0 ^a dt \int _{|x|=t} \frac{f}{|\nabla h|} d\sigma, $$ but note that $\nabla h(x) = x/|x|$, hence $|\nabla h|=1$. $$ \int _0 ^a dt \int _{|x|=t} \frac{f}{|\nabla h|} d\sigma=\int _0 ^a dt \int _{|x|=t} f(x) d\sigma. $$ Use absolute value inequality, $$ \Big| \int_{B_a}f(x)dx \Big| = \Big| \int _0 ^a dt \int _{|x|=t} f(x) d\sigma \Big| \leq \int _0 ^a dt \Big| \int _{|x|=t} f(x) d\sigma \Big|. $$ Then I use (2) to estimate RHS: let $p\in \partial B_{a}$, then by assumption $f(p)=0.$ $$ f(x)=f(x)-f(p)=\int_0 ^1 \nabla f(p+t(x-p))\cdot (x-p)dt, \\ |f(x)| \leq \int_0 ^1 \Big| \nabla f(p+t(x-p))\cdot (x-p) \Big| dt \leq \sup |\nabla f| \cdot |x-p|. $$ If $|x|=t$, then $|f(x)|\leq \sup |\nabla f| \cdot (a+t)$. Hence $$ \int _0 ^a dt \Big| \int _{|x|=t} f(x) d\sigma \Big| \leq \sup |\nabla f| \int _0 ^a dt (a+t) \int _{|x|=t} d\sigma = \\ \sup |\nabla f| \int _0 ^a dt (a+t) \omega_m \cdot t^{m-1}= \\ \frac{2m+1}{m^2+m} a^{m+1} \omega _m \sup_{B_a} |\nabla f|. $$ And it's the best result I could obtain.
Thank you in advance.

Answer 1

Given $x\in B_a$, $x\neq 0$, let $y_x := a x / |x|$, and let $y_0$ be any point of $\partial B_a$, so that $f(y_x) = 0$ for every $x\in B_a$. Denoting $M := \sup_{B_a} |\nabla f|$, we have that $$ f(x) = f(x) - f(y_x) = \int_0^1 \nabla f(y_x + t(x-y_x))\cdot (x-y_x)\, dt $$ so that $$ |f(x)| \leq (a- |x|) M, \qquad \forall x\in B_a. $$ Hence, $$ \int_{B_a} |f(x)|\, dx \leq M \int_{B_a} (a - |x|) \, dx= M \omega_m \frac{a^{m+1}}{m(m+1)}\,. $$