For the heat equation:
$$\frac{\partial u}{\partial t} = \Delta u + \zeta(t,x)$$
where $u:\mathbb{R}^{+}\times \mathbb{R}^{n} \rightarrow \mathbb{R}$ and $\zeta$ is a space-time white noise.
I'm wondering how to understanding the following words:
''We can ‘trade’ space-regularity against time-regularity at a cost of one time derivative for two space derivatives.''
I want an example to understand these words. Could you please show me some examples?
The presence of noise allows to exchange ("trade") time regularity for space regularity. For a simple example, consider the map $$(t,x)\mapsto \int_0^t b(s,x+W_s)\,ds,$$ with Gaussian noise $W_s$. If without the noise the map is $C_t^1C_x^0$, with the noise it becomes $C_t^0 C_x^1$, so the improvement of space regularity is obtained at the cost of loss of time regularity. See these lecture notes (pages 14+15).
In the case of the stochastic heat equation, the noise converts $C_t^{1/2}C_x^0$ regularity into $C_t^{1/4}C_x^{1/2}$ regularity, at a "rate of two powers of space to one power of time". This is derived in section 5.1 of these notes. The heuristics for the two-to-one conversion rate is based on dimensional analysis: The heat equation $u_t=Du_{xx}$ has one dimensionfull coefficient $D$ with dimensions length$^2$/time. This allows the conversion of $p$ powers of space into $q$ powers of time at a ratio $p:q=2:1$.
Indeed for a detailed description see the first seven pages in Hairer SPDE notes. First, lets take a look at the heat equation
$$\partial_{t}u=\Delta u.$$
The idea of "trade" basically says that if we have some expression $G(t,x)=F(u,\partial_{t}u)$ (e.g. that we want to estimate), then we can improve the time-regularity of this expression by replacing it with
$$G(t,x)=F(u,\Delta u).$$
So now our expression $G(t,x)$ improved in time-regularity because we removed $\partial_{t}u$ but it lost two spatial-derivatives i.e. the expression $F(u,\Delta u)$ might have low spatial-regularity due to the insertion of the term $\Delta u$.
Also, from a scaling-perspective, we observe that if $u$ solves the heat equation $u_{t}=\Delta u$, then so does $u^{\lambda}=u(\lambda x,\lambda^{2} t)$. We want a similar invariance for the stochastic heat equation. As mentioned here Scaling of space-time white noise,in order to have
$$\epsilon^{\gamma} \xi(\epsilon^{\alpha}t,\epsilon^{\beta}x)\stackrel{dist}{=}\xi(t,x),$$
we need $\gamma = \alpha \frac{d}{2} + \frac{\beta}{2}$. So by taking $d=1,\alpha=2,\beta=1$, we get $\gamma=\frac{3}{2}$. So that means that $u^{\lambda}=\lambda^{3/2}u(\lambda x,\lambda^{2} t)$ also satisfies the stochastic heat equation.
Therefore, we have the following "trade"
$$\partial_{t}^{\alpha}\leftrightarrow \nabla^{2\alpha}.$$
Now lets return to the stochastic heat equation
$$\partial_{t}u=\Delta u+\xi.$$
Here I try to follow a heuristic on "trading" on page 5 Hairer SPDE notes (any feedback is welcome because the following is my interpretation and I want to understand that formal idea too).
We recall that $E\xi(t,x)\xi(s,y)=\delta(t-s)\delta(x-y)$, meaning that $\xi$ in both t,x is a derivative of Brownian motion and so the regularity will be $\frac{1}{2}-1-\epsilon=-\frac{1}{2}-\epsilon$ in both variables. We want to check whether $u$ will be continuous in both $t,x$.
Here is a very heuristic idea with using inverses. We can "write"
$$\partial_{t}u=\Delta u+\xi\Rightarrow u(t,x)=\partial_{t}^{-1}(\Delta u+\xi).$$
So now this expression $u(t,x)$ is time-regular $C^{1/2-\epsilon}$ but still very irregular in space i.e. $C^{-1/2-\epsilon}$. We can also "write"
$$\Delta u=\partial_{t}u-\xi\Rightarrow u(t,x)=\Delta_{t}^{-1}(\partial_{t}u-\xi).$$
Now $u(t,x)$ is spatially regular $C^{3/2-\epsilon}$ but still very irregular in time i.e. $C^{-1/2-\epsilon}$. So we want to trade between them so that we have simultaneous continuity. We "write"
$$\partial_{t}^{1/2} u(t,x)=\partial_{t}^{-1/2}(\Delta u+\xi).$$
On the RHS, ignoring the $\Delta u$, we have an expression with time-regularity $C^{0}(time)$. Therefore, we no longer have to worry about time-regularity. Thus, if we make the "trade" $\partial_{t}^{1/2}\mapsto \nabla$, we get a gain of $+1-\frac{1}{2}=\frac{1}{2}$ in spatial regularity $$\nabla^{-1}\partial_{t}^{-1/2}(\Delta u+\xi).$$
Similarly, we "write"
$$\nabla^{3/2} u(t,x)=\nabla^{-1/2}(\partial_{t} u-\xi)$$
to get a RHS expression that has spatial-regularity $C^{0}(space)$. Therefore, we no longer have to worry about spatial-regularity. So if we make the trade $\nabla^{3/2}\mapsto \partial_{t}^{3/4}$, we get an expression
$$\partial_{t}^{-3/4}\nabla^{-1/2}(\partial_{t} u-\xi)$$
which has time-regularity $C^{1/4}(space)$.
