Does anyone have any ideas on how to show that the following is true:

Let $\Omega \subset \mathbb{R}^{n}$ be open and bounded.

Consider vector-valued function $$f: \Omega \times \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$$ which has growth conditions $$|f_{i}(x,y,z)| \leq \beta\big(k(x) + |s|^{p_{0}}+\sum_{j=1}^{n}|z_{j}|^{p_{j}}\big)^{1-\frac{1}{p_{i}}}$$ where $\alpha > 0$, $\beta > 0$, $p_{0} > 1$ and $k \in L^{1}(\Omega)$.

Assume $$f(x,y,z)\cdot z \geq \alpha\sum_{i=1}^{n}|z_{i}|^{p_{i}}$$ for a.e. $x \in \Omega$ and every $(y,z)\in \mathbb{R} \times \mathbb{R}^{n}$, where $p=(p_{1},...,p_{n})$ and $p_{i} > 1$. Does the previous condition on $f$ imply the following condition or does the following condition on $f$ imply the previous condition? $$\forall z_{0} \in \mathbb{R}^{n}: \lim\limits_{|z|\rightarrow \infty}\frac{f(x,y,z)\cdot(z-z_{0})}{|z|} = \infty ~~~ for~y~bounded. $$

Thanks for any assistance.

  • 4
    $\begingroup$ Hope you get an answer since you have clearly used most of your reputation for the bounty :) $\endgroup$
    – user100431
    Dec 9, 2014 at 10:36

1 Answer 1


For fixed $x$ and $y$ you have \begin{equation*} f(x,y,z).z\geqslant \alpha \sum |z_{i}|^{p_{i}} \gg |z| \end{equation*} since $|z|\sim \sum |z_{i}|$ (all norms are equivalent in finite dimension), so that indeed $\frac{f(x,y,z).z}{|z|}\rightarrow +\infty$.

Also $f(x,y,z)z_{0}\leqslant C|z_{0}|\sum_{j}|C'+\sum_{i} |z_{i}|^{p_{i}}|^{1-\frac{1}{p_{j}}}$ by Cauchy-Schwarz and equivalence of norms again. This is neglectable when compared to $f(x,y,z).z$ when $z\rightarrow \infty$ because because for every $j$ in the sum, $|C'+\sum_{i} |z_{i}|^{p_{i}}|^{1-\frac{1}{p_{j}}} \ll |\sum_{i} |z_{i}|^{p_{i}}|\leqslant \frac{1}{\alpha}f(x,y,z)z$

Putting all together \begin{equation*} \lim_{z\rightarrow \infty}\frac{f(x,y,z).(z-z_{0})}{|z|}=\frac{f(x,y,z).(z)}{|z|}=\infty \end{equation*}

If this is not what you want perhaps you should precise it in your post. You make a lot of hypothesis concerning $x$ and $y$ that are difficult to understand. In fact saying that the lower bound on $f$ only holds for a.e $x\in \Omega$ (what is $\Omega$ by the way ?) and not on all of $\Omega$ is enough to construct a counter-example to the conclusion you want.


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