How to prove that the delta function is not actually a function

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?

Solution 1:

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.