Take the 2-minute tour ×

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 $\epsilon$

share|improve this question
I believe the "norm" you've given is a seminorm. –  Christopher A. Wong Oct 5 '12 at 0:17
Thanks. You're right, it is a seminorm... We can define a norm in the fashion $$||f||_{C^k} = \sum_{j=0}^k{||D^{j}f||}$$ defined as above. As stated, the question is valid though. (It's just the $C^0$ norm of each term). –  Euler....IS_ALIVE Oct 5 '12 at 0:33
The problem is not the norm, the problem is the domain. This is not a norm nor a seminorm if the domain is unlimited. Is this domain right? Maybe is a bounded domain? –  Tomás Oct 7 '12 at 1:09
Tomas, it is implicit that the norm on the left is finite if both norms on the right are. If one of the norms on the right is infinite, the inequality is trivially true. –  Lukas Geyer Oct 7 '12 at 4:30
But "norms" dont assume the value $\infty$. So this is not a "norm". This is just a extended number involving the derivatives of the function $u$ if the domain is unbounded. –  Tomás Oct 7 '12 at 22:47

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
Reminds me of Littlewood's inequality, which states that if a function and its second derivative are bounded, then its first derivative as a bound depending on these two bounds. –  marty cohen Oct 7 '12 at 1:22
I'm not sure how you get from considering this function to reducing the problem to $n=1$. –  Euler....IS_ALIVE Oct 7 '12 at 3:27
If you denote the original function by $u$ and the function restricted to the line by $f$, then obviously $\|f\| \le \|u\|$, and $|f''(x)| = \| b^t D^2u(x) b \| \le \| D^2u(x) \|_2$ where $\|.\|_2$ is the spectral norm of the matrix. Since any two matrix norms are comparable and your norm is the max-norm of $D^2u$ (and you can certainly find the exact constant easily), we have $\|f\| \le K_n \|Df\|$. Now if you know the result for $n=1$, you can conclude that your inequality holds for $f'$, the directional derivative of $u$. It holds for all of those, so it holds for the gradient. –  Lukas Geyer Oct 7 '12 at 4:28
......What? Shouldn't the proof depend on Taylor's theorem or something like that? Where are matrices coming in? And did you switch from $t$ to $x$? And what exactly do you mean by $f''$, and how did you get that equality in your comment. I seem to be very confused. –  Euler....IS_ALIVE Oct 7 '12 at 5:55
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. –  Lukas Geyer Oct 7 '12 at 14:02

Your Answer


