Fourier transform of the 1-dimension heat equation For the one dimentional heat equation $u_t=u_{xx}$, the book i am reading takes fourier transform on both sides of the equation with respect to x. My question is the following. 
1: Is the fourier transform of $u_{xx}$ equal to the second derivative of U, the fourier transform of u? i i.e. is it true that $F(u_{xx}) = F(u)_{xx}$ (F is fourier operator)? If not, which is the right approach in this case?
2: why is the right-hand-side, after taking the fourier transform, equal to $-4\pi^2f^2U(f, t)$?
 A: There are assumptions that must be placed on $u$ to even make sense of the Fourier transform. For the heat equation, it makes the most sense to assume $u$ is absolutely integrable on $\mathbb{R}$ in order for the total heat to be finite. However, you could instead assume $u$ is square integrable, even though this is not necessarily physical for the heat equation. If you assume $u$ is locally absolutely continuous in $x$, with locally absolutely continuous $u_x$, and if you assume $u,u_x,u_{xx}$ are absolutely square integrable, then the integration by parts evaluation terms vanish and you get what you want. For example,
$$
      (\mathcal{F}u_{x})(s,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u_x(x,t)e^{-isx}dx \\
  = \frac{1}{\sqrt{2\pi}}u(x,t)e^{-isx}|_{x=-\infty}^{\infty}+\frac{is}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u(x,t)e^{-isx}dx \\
  = (is\mathcal{F}u)(s,t).
$$
Either set of assumptions mentioned above will allow you to conclude that the integration-by-parts evaluation terms vanish, leaving
$$
                \mathcal{F}u_x = is\mathcal{F}u \\
         \mathcal{F}u_{xx} = is\mathcal{F}u_x = -s^2\mathcal{F}u.
$$
However, what is not obvious is an assumption that will allow an interchange of the time derivative with the Fourier integral
$$
            (\mathcal{F}u_t) = (\mathcal{F}u)_t,
$$
which is to say
$$
     \lim_{h\downarrow 0}\mathcal{F}\left(\frac{u(\cdot,t+h)-u(\cdot,t)}{h}\right)=\mathcal{F}(u_t).
$$
A typical assumption is that the difference quotient converges as $h\downarrow 0$ to $u_t$ either in $L^1$ or in $L^2$. If this holds, then
$$
          (\mathcal{F}u)_t = -s^2\mathcal{F}u \\
      \implies (\mathcal{F}u)(s,t) = (\mathcal{F}u)|_{t=0}e^{-s^2t}.
$$
A: The reason why the fourier transform is used in PDE is that it has very nice properties with respect to differentiation.
Let $\mathcal{F} (f) $ denote the Fourier transform of $f$ then you can easily get the Fourier transform of $f'$ from it: 
$$ \mathcal{F}(f') (\xi) = 2 \pi i \xi \mathcal{F} (f) (\xi) $$  
Apply this again to obtain the Fourier  transform of the second derivative
$$ \mathcal{F}(f'') (\xi) = (2 \pi i \xi )^{2} \mathcal{F}(f)(\xi) $$
This is for a real function $f$, in your case you have a function of two variables $(t,x)$. 
Now, try to compute the Fourier transform of $u$. 
$$ \mathcal{F}(u) (\omega) = \int_{\mathbb{R} \times \mathbb{R}} u(t,x)e^{-2 \pi i(t \omega_1 + x \omega_2 )} dx dt $$ 
where of course $ \omega = (\omega_1,\omega_2) $. 
Now you can return to your problem, you can explicitly compute the Fourier transform of $ u_t $ and $ u_{xx} $. (Hint: Separate the integral into $2$ integrate one first....) 
Can you finish from here? 
