How to find the coordinate representation of the kinetic operator? From my professor's notes on statistical mechanics. $\left|\bf{k}\right\rangle$ is eigenstate of the hamiltonian of the free particle with periodic boundary conditions:
$$
\left\langle{\bf r}|{\bf k}\right\rangle = \frac{1}{\sqrt{V}} e^{i\bf{k}\cdot\bf{r}}.
$$

The states $\left|\bf{k}\right\rangle$ are eigenstates of the kinetic
  energy: $T_1|{\bf k}\rangle = (\hbar^2
 k^2/2m)\left|\bf{k}\right\rangle$. Thus, the single-particle kinetic
  energy is diagonal in momentum space $$ \langle{\bf k'} | T_1 | {\bf
k}\rangle = \delta_{{\bf k},{\bf k'}}\frac{\hbar^2 k^2}{2m}. $$
One gets \begin{align} \delta_{{\bf k},{\bf k'}}\frac{\hbar^2 k^2}{2m} &= \int d{\bf r'} \int d{\bf r} \langle{\bf k'}|{\bf r'}\rangle
\langle{\bf r'} | T_1 | {\bf r}\rangle \langle{\bf r}|{\bf k}\rangle
\\ &= \int d{\bf r'} \int d{\bf r} \frac{1}{\sqrt{V}} e^{-i\bf{k'}\cdot\bf{r'}} \langle{\bf r'} | T_1 | {\bf r}\rangle
\frac{1}{\sqrt{V}} e^{i\bf{k}\cdot\bf{r}} \end{align} from which we
  have the well-known coordinate representation of the kinetic-energy
  operator $$ \langle{\bf r'} | T_1 | {\bf r}\rangle = \left(\frac{-\hbar^2 \nabla^2}{2m}\right)\delta({\bf r}-{\bf r'}). $$

What I don't understand is how it concludes the very last equation from the one preceeding it. I could verify it is right by substituting it back, but I'm interested in another explanation and more importantly its uniqueness. I think there is a trivial way to say that, but I couldn't find it. Any help is greatly appreciated. 
 A: I've relabeled $\bf r,\, r'$ from the initial integral to:
$$ \delta_{\bf k,k'} \frac{\hbar^2k^2}{2m} = \frac{1}{V}\int\mathrm{d}{\bf r'_1}\int\mathrm{d}{\bf r_1}e^{-i\bf k' \cdot r'_1}\langle{\bf r'_1}|T_1|{\bf r_1}\rangle e^{i\bf k \cdot r_1} $$
As ZachMcDargh mentions in his answer, the following identity is useful:
$$ \sum_{\bf k} e^{-i\bf k \cdot r} = V \delta(\bf r) $$
Taking two (discrete) inverse Fourier transforms on both sides:
\begin{align} \sum_{\bf k'}\sum_{\bf k}e^{i\bf k' \cdot r'}\delta_{\bf k,k'} \frac{\hbar^2k^2}{2m}e^{-i\bf k \cdot r} & = \frac{1}{V}\sum_{\bf k'}\sum_{\bf k}\int\mathrm{d}{\bf r'_1}\int\mathrm{d}{\bf r_1}\;e^{-i\bf k' \cdot (r'_1 - r')}\langle{\bf r'_1}|T_1|{\bf r_1}\rangle e^{i\bf k \cdot (r_1 - r)} \\
 & = V\int\mathrm{d}{\bf r'_1}\int\mathrm{d}{\bf r_1}\;\delta({\bf r_1' - r'})\delta({\bf r_1 - r})\langle{\bf r'_1}|T_1|{\bf r_1}\rangle \\
\implies \sum_{\bf k} \frac{1}{V}\frac{\hbar^2k^2}{2m}e^{-i\bf k \cdot (r-r')} & = \langle{\bf r'}|T_1|{\bf r}\rangle \end{align}
Evaluate the sum on the LHS in Cartesian coordinates:
$$ \implies \langle{\bf r'}|T_1|{\bf r}\rangle = \frac{\hbar^2}{2mV}\sum_{\bf k} [k_x^2 + k_y^2 + k_z^2]\;e^{-i\bf k \cdot (r-r')}$$
Note that (via partial differentiation under the integral sign, as long as $\bf r$ is differentiable):
$$ \sum_{\bf k}\; k_m^2 \;e^{-i\bf k \cdot (r-r')} = -\frac{\partial^2}{\partial m^2}\sum_{\bf k}\;e^{-i\bf k \cdot (r-r')},\quad m = x,\,y,\,z$$
Therefore:
\begin{align} \langle{\bf r'}|T_1|{\bf r}\rangle & = -\frac{\hbar^2}{2mV}\sum_{\bf k} \; \left[\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial x^2}\right]e^{-i\bf k \cdot (r-r')} \\
& = -\frac{\hbar^2 \nabla^2}{2mV}\sum_{\bf k}\;e^{-i\bf k \cdot (r-r')} = -\frac{\hbar^2 \nabla^2}{2m} \delta(\bf r - r') \end{align}
The same as the result noted by AccidentalFourierTransform (not so accidental in this answer, heh). 
Unfortunately I don't know distribution theory, so I can't prove uniqueness yet.
A: In order to derive the last equation from the one before it, multiply both sides by $e^{-i \boldsymbol{k}\cdot \boldsymbol{r}''}e^{i \boldsymbol{k}'\cdot \boldsymbol{r}'''}$, and sum over both $\boldsymbol{k}$ and $\boldsymbol{k}'$, i.e. perform inverse Fourier transforms. Using the identity
$$
\sum_{\boldsymbol{k}} e^{i \boldsymbol{k}\cdot\boldsymbol{r}}= V\delta(\boldsymbol{r})
$$
you'll quickly see that 
$$
\langle \boldsymbol{r}' | T_1  | \boldsymbol{r}\rangle = 
\sum_{\boldsymbol{k}} \frac{1}{V}
e^{-i \boldsymbol{k} \cdot(\boldsymbol{r} - \boldsymbol{r}')}
\frac{\hbar^2 k^2}{2 m}.
$$
Then we just identify the Fourier transform of $k^2$ as $-V\delta(\boldsymbol{r}-\boldsymbol{r}')\nabla^2$.
