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Divergence of Fourier transforms of a tempered distribution

I am having trouble with the following:
Let $a\in\mathbb{R}$ such that $|a|>0$. Let $\phi\in S'(\mathbb{R})$ with $\widehat{\phi}(0)=1$ and $\widehat{\phi}\geq0$ on $[-1,1]$, so that $\phi \in W^{1,\infty}([-1,1])$.
I want to show that
$$
\int_{ -\infty}^\infty \frac{\chi_{[-\epsilon,\epsilon]}(x)}{x}\phi(x)\mathrm{d}x=c\epsilon,
$$
where $c$ is some constant. The identity (which I have already shown)
$$
\int_{ -\infty}^\infty \frac{\chi_{[-\epsilon,\epsilon]}(x)}{x}\mathrm{d}x=2\epsilon
$$
is clearly not useful to me. I’ve tried looking up Fourier transforms of $\chi_{[-\epsilon,\epsilon]}$ and trying to convolve them with $\phi$, but I am getting nowhere. Any help would be appreciated!

A:

To prove your integral is equal to $2\epsilon$ you’ll have to take things apart more carefully.
With what you’ve told

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Q:

How to prove this inequality? $3x^3+6x^2+2x+1\geq6x^2+9x+3$

How to prove this inequality? $3x^3+6x^2+2x+1\geq6x^2+9x+3$
I tried to prove it by AM-GM. I cannot get the best lower bound.

A:

Start by writing it out as:
$$3x^3+6x^2+2x+1-6x^2-9x-3=x(3x^2+2x-6)$$
Now, use AM-GM:
$$\frac{3x^2+2x-6}{3}\times\frac{3x^2+2x-6}{3}\geq\frac{3(3x^2+2x-6)^2}{9}=\frac{3\cdot18}{3}=6$$
Again, using AM-GM:
$$\frac{6x^2+9x+3}{6}\times\frac{6x^2+9x+3}{6}\geq\frac{(6x^2+9x+3)(6x^2+9x+3)}{36}=\frac{81x^4+108x^3+81x^2+18x+9}{36}\geq\frac{81x^4

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