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Let $f\in C^2(R^m), f\geq 0$$f\in C^2(\Bbb R^m), f\geq 0$, Hessian matrix of $f$ is upper bounded by some constant $C$. Do we have $|\nabla f|\leq \alpha \sqrt{f}$ for some $\alpha$, even if the Hessian matrix is degenerate?
a property of C^2 functions
Let $f\in C^2(R^m), f\geq 0$, Hessian matrix of $f$ is upper bounded by some constant $C$. Do we have $|\nabla f|\leq \alpha \sqrt{f}$ for some $\alpha$, even if the Hessian matrix is degenerate?
A property of $C^2$ functions
Let $f\in C^2(\Bbb R^m), f\geq 0$, Hessian matrix of $f$ is upper bounded by some constant $C$. Do we have $|\nabla f|\leq \alpha \sqrt{f}$ for some $\alpha$, even if the Hessian matrix is degenerate?
Let $f\in C^2(R^m), f\leq 0$$f\in C^2(R^m), f\geq 0$, Hessian matrix of $f$ is upper bounded by some constant $C$. Do we have $|\nabla f|\leq \alpha \sqrt{f}$ for some $\alpha$, even if the Hessian matrix is degenerate?
Let $f\in C^2(R^m), f\leq 0$, Hessian matrix of $f$ is upper bounded by some constant $C$. Do we have $|\nabla f|\leq \alpha \sqrt{f}$ for some $\alpha$, even if the Hessian matrix is degenerate?
Let $f\in C^2(R^m), f\geq 0$, Hessian matrix of $f$ is upper bounded by some constant $C$. Do we have $|\nabla f|\leq \alpha \sqrt{f}$ for some $\alpha$, even if the Hessian matrix is degenerate?
