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How can I show that there cannot be a function $\delta$ such that $\int_{-1}^1f(x)\delta(x)\,dx = f(0)$ (in particular $f(x)\delta(x)$ is Riemann integrable on $[-1, 1]$) for every $f$ continuous on $[-1,1]$?

So far I've figured out that if $\delta$ were to exist, then $\int_{-1}^1\delta(x)\,dx = 1$ and $\int_{-1}^1f(x)\delta(x)\,dx= \int_{-1}^1f(0)\delta(x)\,dx$ for all $f$. Can I somehow produce a contradiction (for example, show that $\delta$ is unbounded, or show that it is zero everywhere except the origin, etc.)? Is there some other fact about $\delta$ that I can deduce which will lead me to a contradiction?

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    I think it's just not a function by definition, but I'm probably wrong. – 2017-01-25

2 Answers 2

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We must not assume that $\delta$ is continuous. However, it must be Riemann-integrable: Therefore $\delta$ must be bounded near $0$, say $|\delta(x)|0$, consider $f(x)=\max\{0,1-|x/r|\}$. Then $f$ is a continuous function on $[-1,1]$ with $0\le f(x)\le 1$ for all $x$, $f(x)=0$ for $|x|>r$, and $f(1)=1$. It follows that for $r<\frac12$ we have $$1=\left|\int_{-1}^1f(x)\delta(x)\,\mathrm dx \right|\le \int_{-1}^1\left|f(x)\delta(x)\right|\,\mathrm dx\le \int_{-r}^r M\,\mathrm dx=2Mr$$ If $r<\frac1{2M}$, this is a contradiction.

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    Is there any reason you use $1/2$? Could we not say $\lvert \delta(x) \rvert < M$ for all $x \in [-1,1]$? Because $\delta$ should be bounded on the whole interval? – 2017-01-22
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Yes, you can see that δ is unbounded.

See $f_n = max(0, 1-nx^2)$ with n -> $\infty$

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    "can see what" should be "can see that"? It would substantially improve your Answer to spell out how the defined sequence shows that $\delta$ is unbounded (as it would also be good to explain in what way this prevents $\delta$ from being "a real function" as the Question seems to ask). – 2017-01-22