Let $M$ be a smooth compact riemannian manifold with Levi Civita connection and consider a smooth function $f: M \to \mathbb{R}$. Then the Laplacian of $f$ $$ \Delta f = \text{div} ( \text{grad} f) $$ end the Hessian of $f$ $$ \nabla^2 f (X,Y) = \langle \nabla_X \text{grad} f , Y \rangle $$ are well defined. I want to prove the inequality $$ \| \Delta f\|_{\infty} \le C\| \nabla^2 f \|$$ for some positive constant $C$. I'm using $$ \| \Delta f\|_{\infty} = \sup_{x \in M} |\Delta f| \quad \quad \|\nabla^2 f\|_{\infty} = \sup_{x \in M} \Biggl ( \sup_{u,v \in T_xM \setminus \{0\}} \frac{ |\nabla^2 f (u,v)|}{|u||v|} \Biggr ) $$
How can I prove it?
Looking at the problem in an abstract way, maybe I can consider a general $\tau \in T_2^0(M)$ and since the laplacian is the trace of the hessian I have
$$ \|\text{tr}(\tau)\|_{\infty} = \sup_{x \in M} |\text{tr}(\tau(x))|= \sup_{x \in M} \tau_{ij}(x)g^{ij}(x) \le \sup_{x \in M} \biggl ( \sum_{i,j}\tau_{ij}(x) \max_{ij} g^{ij}(x) \biggr ) =$$ $$ \sup_{x \in M} \biggl ( \max_{ij}g^{ij}(x) \sum_{i,j} \tau_{ij}(x) \biggr ) = \biggl ( \sup_{x \in M} \max_{i,j} g^{ij}(x) \biggr ) \sup_{x \in M} \sum_{i,j}\tau_{ij}(x) \lesssim \biggl ( \sup_{x \in M} \max_{i,j} g^{ij}(x) \biggr ) \|\tau\|_{\infty}$$ where the last inequality is due to the fact that all norms are equivalent in finite dimensional spaces. Moreover the quantity $$ \biggl ( \sup_{x \in M} \max_{i,j} g^{ij}(x) \biggr )$$ is finite being it the supremum of a continuous function on a compact set.