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Lef $u$ be at least a $C^2$ function on $\mathbb{R}^n$. Let's denote the gradient by $D$. Also, (using the multiindex notation), define the seminorm $||D^ku|| = \sup_{|\gamma|=k}{\sup_x{|D^{\gamma}u|}}$

How can we prove the following : $||Du|| \leq \epsilon||D^2u|| + C||u|| $ where $C$ is some constant depending on $\epsilon$

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    If $\Omega$ is bounded, thats what im saying, but in the statement of the problem you write: Let $u$ be at least a $C^{2}$ function in "$\mathbb{R}^{n}$". $\mathbb{R}^{n}$ is not bounded, so this is not a norm nor a seminorm.2012-10-08

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By restricting the function to a line, i.e., considering the function, $t \mapsto u(a+tb)$ for some point $a\in \mathbb{R}^n$ and a unit vector $b\in\mathbb{R}^n$, you can reduce the problem to the case $n=1$. Now the problem is for a $\mathcal{C}^2$ function $f:\mathbb{R}\to\mathbb{R}$ to show $\|f'\| \le \epsilon \|f''\| + C\|f\|$. The idea is that if you have a point $x_0$ and a constant $M>0$ such that $f'(x_0)\ge M+1$ (or $-f'(x_0)\ge M+1$), and if you have a uniform bound $\|f''\|\le K$, then $f'\ge M$ (or $-f'\ge M$) whenever $|x-x_0| \le 1/K$, and then $|f(x_0+1/K)-f(x_0-1/K)| \ge 2M/K$, so $\| f\| \ge M/K$. This implies the desired inequality by juggling of constants.

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    Matrices are coming in when using the chain rule for functions of several variables. The $b^t$ is the transpose of $b$ (i.e., a row vector instead of a column vector), and $f''$ is the second derivative of f. The equality is exactly the chain rule for the second derivative of $f$ as defined. You don't really need the matrix notation and the spectral norm here, but it makes the notation more compact. The important part is that you get a norm estimate for $f''$ in terms of the bound for $D^2u$, and you can get that from whatever form of the chain rule you know.2012-10-07