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Here is another Prelim Question.

Evaluate: $$ I = \lim_{\epsilon\to 0} \frac{1}{2\pi\epsilon}\int_{-\infty}^\infty\int_{-\infty}^\infty f(x,y)\frac{y}{\epsilon}\exp{\frac{-(x^2+y^2)}{2\epsilon}}dxdy $$ where $f$ and $\nabla f$ are continuous and bounded.

So I have tried to change the integration order and used integration by parts and then changing to polar coordinates to obtain $$ I = \frac{\partial f}{\partial y}(0,0) + \lim_{\epsilon \to 0}\frac{1}{2 \pi}\int_0^{2\pi}\int_0^\infty\frac{\partial}{\partial r}\left(\frac{\partial f}{\partial y}(r\cos(\theta),r\sin(\theta) )\right) \exp\left(\frac{-r^2}{2\epsilon}\right) dr d\theta $$ I hope the final answer is the first term and we can show the second term goes to zero. Can we just apply DCT? I am getting second order partials in the second term and we only know information for the first order derivatives. Note that $$ \frac{\partial }{\partial r} = \cos(\theta)\frac{\partial}{\partial x}+\sin(\theta)\frac{\partial}{\partial y} $$

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Integrate w.r.t. $y$ first and use integration by parts to get $$ I = \frac{1}{2\pi\epsilon}\int_{-\infty}^\infty\int_{-\infty}^\infty f_y(x,y)\exp{\frac{-(x^2+y^2)}{2\epsilon}}dydx $$ Here you have used the fact that $$\lim_{y\to\pm\infty} f(x,y)\exp{\frac{-(x^2+y^2)}{2\epsilon}}=0$$ for each fixed $x$ and that $$\frac{\partial}{\partial y} \exp{\frac{-(x^2+y^2)}{2\epsilon}}=-\frac{y}{\epsilon}\exp{\frac{-(x^2+y^2)}{2\epsilon}}$$ Now $$\frac{1}{2\pi\epsilon}\int_{-\infty}^\infty\int_{-\infty}^\infty \exp{\frac{-(x^2+y^2)}{2\epsilon}}dydx=\frac{1}{2\pi\epsilon}\cdot (2\pi)\int_0^\infty\exp(\frac{-r^2}{2\epsilon})r\,dr=1$$ for every $\epsilon>0$. Since the function $\frac{1}{2\pi\epsilon}\exp{\frac{-(x^2+y^2)}{2\epsilon}}$ is nearly support at $(0,0)$ for small $\epsilon$, you have an approximation to the delta function. Hence your required limit is $f_y(0,0)$.

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