1
$\begingroup$

Suppose $A$, $B$, $C$ and $D$ are fixed arbitrary positive numbers.

I am free to choose $\epsilon$, $\epsilon_1$ and $\epsilon_2$ but they must be positive.

Can I choose the epsilons so that $\left((A -\frac{1}{\epsilon_1} - \frac{B}{2\epsilon_2})\frac{C\epsilon}{(1+\epsilon)} - \epsilon_1 D\right) > 0?$

I need this to show coercivity of a bilinear form.. Thanks.

I did some calculations, and it seems if I make $\epsilon_1$ very small and $\frac{\epsilon}{1+\epsilon}$ small then it might work but I can't prove it. Anyway I really hope it can be done.

Thanks

  • 0
    @Hurkyl I didn't try that.. I wanted to avoid the differentiation. But I will give it a go.2012-12-24

1 Answers 1

3

Let $\alpha=\frac{1}{\epsilon_1}$, $\beta=\frac{1}{2\epsilon_2}$ and $\gamma=\frac{\epsilon}{1+\epsilon}$. Then you want to find a triple $(\alpha,\beta,\gamma)$ with $\alpha,\beta>0$ and $0<\gamma<1$ and, after dividing both sides by $\epsilon_1$ and rearranging, the condition:

$\alpha(A-\alpha-B\beta)C\gamma > D$

In particular, you need:

$\alpha(A-\alpha-B\beta)> D/C$

and if you do have this, you can easily find $\gamma$.

Now, the supremum of $\alpha(A-\alpha-B\beta)$ is the supremum of $\alpha(A-\alpha)$, which is when $\alpha=A/2$, and yields the value $\frac{A^2}4$. So at minimum, you need:

$\frac{A^2}{4}> D/C$

You can pick $(\alpha,\beta,\gamma)$ (and hence $\epsilon,\epsilon_1,\epsilon_2$) if and ony if this is true.

  • 0
    That's a shame, thanks.2012-12-24