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Show that there is no continuous bounded function $\delta : [−1/2, 1/2) → R$ with the following property: for all continuous bounded functions $f : [−1/2, 1/2) → R$, $$\int_{-1/2}^{1/2}f(x)\delta(x)=f(0)$$

The point of this is to show that the delta function is not actually a function so we need to generalize functions into distributions. Given that $0$ is an arbitrary point in the interval, this statement is saying that the integral of the product of two functions cannot just happen to be the first function evaluated at some point. I was surprised that this would be the case since $\delta(x)$ can take on negative values, I thought it might be possible that the negative and positive parts can sum up to $f(0)$

As for how to prove this result, I thought this would contradict some mean value theorem of the fundamental theorem of calculus, but that didn't really work because I can't isolate f(x) by itself. $\delta(x)$ would still show up. A direct proof seems way harder, so proof by contradiction should be the way to go right?

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    $\begingroup$ Consider a family of continuous functions $f_n(x)$, such that $f_n(0)=1$ and $f_n(x)=0$ if $|x|≥\frac 1n$. (easiest, perhaps, to imagine a family of isosceles triangles with shrinking base). $\endgroup$
    – lulu
    Jan 4 at 17:01
  • $\begingroup$ Have you seen the Cauchy-Schwarz inequality? $\endgroup$
    – Ian
    Jan 4 at 17:08
  • $\begingroup$ @Ian, i guess we can impose an inner product structure as the integral from -1/2 to 1/2 which would give us the left hand side. But what about the right hand side? $\endgroup$
    – Bill
    Jan 4 at 18:38
  • $\begingroup$ $g(x)\to\delta(x)$ if $\forall\,f(x)\in C^\infty_c(\Bbb{R}), \int_{-\infty}^\infty g(x)f(x)dx\to f(0)$. For example consider the following limit representation of $\delta (x)$: $g(x)=\underset{f\to \infty}{\text{lim}}\frac{\sin (2 \pi f x)}{\pi x}$ which leads to the following limit representation of $\theta(x)$: $\underset{f\to \infty }{\text{lim}}\left(\frac{1}{2}+\frac{\text{Si}(2 \pi f x)}{\pi }\right)$. $\endgroup$
    – Steven Clark
    Jan 4 at 19:01
  • $\begingroup$ @Bill The idea of the Cauchy-Schwarz argument is that if $\| f_n \|_{L^2} \to 0$ but $|f_n(0)|=|\langle f_n,\delta \rangle_{L^2}| \not \to 0$, then you conclude that $\| \delta \|_{L^2}$ must be arbitrarily large. One can achieve this using an example like lulu's. $\endgroup$
    – Ian
    Jan 4 at 19:05

1 Answer 1

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Assume such a function $\delta$ exists. Following @lulu's advice, define $f_n$ on $(-\frac 1 2, \frac 1 2)$ such that $$f_n(x)=\left \{\begin{array}{ccc} n(x+\frac 1 n) & \text{ if } -\frac 1 n \leq x \leq 0\\ n(\frac 1 n-x) & \text{ if } 0\leq x \leq\frac 1 n \\ 0 & \text{ otherwise} \end{array}\right.$$ Then $f_n$ is continuous, $f_n(0)=1=\max|f_n|$. Moreover $$1=f(0)=\int_{-\frac 1 2}^{\frac 1 2}f_n(x)\delta(x)dx\leq \int_{-\frac 1 {2n}}^{\frac 1 {2n}}|f_n(x)\delta(x)|dx\leq \frac 1 n\max_{[-\frac 1 2, \frac 1 2]} |\delta|$$ and if $n\rightarrow +\infty$, the right-hand side converges to $0$, which is impossible.

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