Here is my problem, as written:
A Dirac approximation sequence is a family of smooth functions $\{ \delta_1, \delta_2, \dots\}$ defined over $\mathbb R$ such that $\delta_n(x) = 0$ whenever $|x| > 1/n$, and for all $\epsilon$, there exists $N$ such that $\int_{-\epsilon}^{\epsilon} \delta_n(x)dx= 1 $ whenever $n>N$. Prove the sifting property of the Dirac delta: that if $f:\mathbb R\to\mathbb R$ is continuous, then $\displaystyle\lim_{n\to \infty} \int_{-\infty}^\infty \delta_n(x)f(x) dx = f(0).$
Here is a sketch of my approach: assume $f(0)=0$ by shifting. Now given fixed $\epsilon$, and for some $N$, we have for large $n>N$ that $$ \left|\int_{-\infty}^{\infty} \delta_n(x) f(x) \right| = \left| \int_{-1/N}^{1/N} \delta_n(x) f(x)\right|.$$ Now since $f$ is continuous at $0$, if $N$ is chosen large enough, I can bound $|f|$ in $[-1/N,1/N]$ and find that the above is at most, say, $$ \left|\int_{-\infty}^{\infty} \delta_n(x) f(x) \right| \leq \epsilon' \left| \int_{-1/N}^{1/N} \delta_n(x)\right|$$ where $\epsilon'$ is my chosen bound for $f$. So I need to bound the right expression with something independent of $n$: this could be done if $\delta$ is presumed to be non-negative (which, as far as I can tell, is how this question is answered elsewhere on this site). Presumably, I need to use the $\int \delta_n = 1$ property, but how can I continue without this non-negativity assumption?
Without non-negativity, it doesn't have to be true.
Let $\omega(x)$ be bump function: smooth, symmetric around $0$, $\omega(x) = 1$ when $|x| < 1/3$, $\omega(x) = 0$ when $|x| \geq 1$, $|\omega| \leq 1$, $\int\limits_{-1}^1 \omega(x)\, dx = 1$. Let $\delta_n(x) = \omega(n(x + 1/n)) \cdot n^2 - \omega(n(x - 1/n)) \cdot n^2 + \omega(nx) \cdot n$ - bumps around $-1/n$ and $1/n$ that compensate each other in integral, plus bump around $0$. Let $f(x) = x$. We have
$$\int_\mathbb R \delta_n(x) f(x)\, dx = \\ \int_{-2/n}^0 \omega(n(x + 1/n))\cdot n^2 \cdot x\, dx - \int_{0}^{2/n}\omega(n(x - 1/n))\cdot n^2 \cdot x\, dx + \int_{-1/n}^{1/n}\omega(nx) \cdot x\, dx = \\ -2\int_{0}^{2/n}\omega(n(x - 1/n))\cdot n^2 \cdot x\, dx < \\ -2\int_{2/(3n)}^{4/(3n)} 1 \cdot n^2 \cdot x\, dx < \frac{-2n}{3} \cdot \frac{2}{3n} = \frac{-4}{9} $$
The problem is, even if $f$ is continuous, it can still break "balance" between parts that "compensated" each other in $\delta_n$ - in this case, by changing sign. Yes, for it $f$ has to be small near $0$, but the parts that "compensated" each other can be taken large enough so that even after multiplication by small value, their integral stays bounded away from $0$.