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I want to see if there exists an $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for every $\epsilon > 0,$ $$\int^{\epsilon}_{-2\epsilon} f(x) dx \geq 1.$$ This seemed like a difficult question, and so I went about playing with the properties of certain functions.

I need to find some function $f$ on the real number line that is continuous and for every $\epsilon > 0$. $g(\epsilon) - g(-2\epsilon) \geq 1,$ where $g'(x) = f(x)$ for all $x$. It would be convenient to assume that $g$ is linear, i.e. set $$g(\epsilon) - g(-2\epsilon) = g(\epsilon) + 2g(\epsilon) = 3g(\epsilon) \geq 1$$ $$\implies g(\epsilon) \geq \frac{1}{3}.$$ The problem I am having is every time I seem to construct a $g$ that satisfies this problem, it is a $g$ that is contingent on some constant, which would go to $0$ when we attempt to find $g' = f$. Is it reasonable to use this mechanism to solve this problem? If so, are there any any suggestions for what to do next?

Sarah
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3 Answers3

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For the integral to be well defined you need at least that the function you seek is Lebesgue integrable. So assuming that, there is a nice theorem which states that if the measure of integration - in your case $3\epsilon$ - goes to zero then the integral will go to zero as well. So there is no such function in the set of Lebesgue measurable functions.

Of course, the named $\delta$ distribution satisfies what you ask but it is not a function...

Alex
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There is no such continuous function $f$ so that $$ \int_{-2\epsilon}^\epsilon f(x) dx \geq 1$$ for all $\epsilon > 0$.

If there were, then $f$ is continuous on $[-1, 1]$. As this is a compact interval, $f$ takes on a maximum $M$ and a minimum $m$, each finite. But then $$ \int_{-2\epsilon}^\epsilon f(x) dx \leq 3\epsilon M,$$ which is clearly less than $1$ for sufficiently small $\epsilon$.

If we remove the necessity for $f$ to be continuous, then we can create some silly functions which satisfy the property. For instance, consider $f(x) = \frac{1}{\lvert x \rvert}$, except at $x = 0$, when we define $f(0) = 0$. Then clearly $$ \int_{-2\epsilon}^{\epsilon} f(x) dx = \infty,$$ which one might think of as bigger than $1$. [But then the function isn't really integrable].

davidlowryduda
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As pointed out by mixedmath, there cannot be a continuous function that satisfies this condition, and as pointed out by Alex there is no Lebesgue integrable function. However, if we expand the acceptable functions and the sense of integration we can cook up something other than the Dirac delta function.

Using the Cauchy Principal Value, we can set $f(x)=-\frac2x$. Then $$ \begin{align} \mathrm{PV}\int_{-2\epsilon}^\epsilon f(x)\,\mathrm{d}x &=\int_{-2\epsilon}^{-\epsilon} f(x)\,\mathrm{d}x +\mathrm{PV}\int_{-\epsilon}^{\epsilon} f(x)\,\mathrm{d}x\\ &=2\log(2)\\[6pt] &\ge1 \end{align} $$ Although integrable in the principal value sense, $-\frac2x$ is not integrable in the general sense.

robjohn
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