I want to say the following: $$\int_0^1 \delta(x-y)f(y)\,dy = f(x)$$ where $f\in C^0([0,1])$. However, I would like to do this in the distributional sense.
As I understand it, for a given distribution $\phi(x,y)\in\mathcal{D}'$ and a bounded region $\Omega$, the integral $\int_\Omega \phi(x,y) dy$ is defined by $$\Big(\int_\Omega \phi(x,y)dy, \varphi(x) \Big) \equiv \lim_{k\rightarrow\infty} \Big(\phi(x,y),\varphi(x)\eta_j(y) \Big)$$ where $\eta_k$ is any test function that is $0$ outside of $\Omega$ and $1$ inside $\Omega$ except within a distance $\sigma(j)$ of the boundary and $\sigma(j)\rightarrow 0$ as $j\rightarrow\infty$.
Since we can only multiply distributions by $C^\infty$ functions, let $\{\psi_k\}$ be a sequence of test functions such that $\psi_k\rightarrow f$. Then $g_k(x,y)\equiv \delta(x-y)\psi_k(y)$ defines a distribution. Then, $$ \Big(\int_0^1 g_k(x,y)dy, \varphi(x) \Big) = \lim_{j\rightarrow\infty} \Big(\delta(x-y)\psi_k(y), \varphi(x)\eta_j(y) \Big)$$ where $\eta_j$ is now over the region $[0,1]$. Now this is where I get stuck, because it seems \begin{align*} \lim_{j\rightarrow\infty} \Big(\delta(x-y)\psi_k(y), \varphi(x)\eta_j(y) \Big) &= \lim_{j\rightarrow\infty} \Big( \delta(x-y) , \psi_k(y) \varphi(x)\eta_j(y) \Big) \\ &= \lim_{j\rightarrow\infty} \psi_k(x)\varphi(x)\eta_j(x) \\ &= \psi_k(x)\varphi(x),\end{align*} and in the limit $k\rightarrow\infty$ this is $f(x)\varphi(x)$. This is not exactly the $$\int f(x)\varphi(x)dx = \Big(f(x),\varphi(x)\Big)$$ that I want. I think I am likely mistaken somewhere and any help would be appreciated.