Can you be more specific in the norm of the derivative of a convolution and Cauchy Schwarz being applied to it? I don't understand how these $L^2$ norms appear, I think I don't have background on these things

@LucasZanella I've added some lines, does it help?

what do you mean when you write $L_x^2$ for example? What is the sub $x$ for?

@LucasZanella It is to emphasise that the integral defining the norm is taken in the $x$ variable. e.g. if $y,z$ is fixed and $g(x) = f(x,y,z)$ then $$ \|f(x,y,z)\|_{L^p_x} := \|g\|_{L^p}$$ ps - If someone has better notation that does not involve replacing variables with dots or $\mapsto$ or defining another function (the first two are horrendously ugly) I'm all ears

Thanks, I understoodd. But now what does $L^2$ alone mean? In which variable?

@LucasZanella In this answer, if I didn't subscript, then its always in the spatial variable (and in each one there is a unique spatial variable so there is no ambiguity). Which line are you talking about?

From line $4$ to $5$ you go from $L_y^2$ to $L^2$

I was in doubt because $t$ can also vary so I could take the norm in $t$, but apparently you made in $x$

@LucasZanella this is the translation symmetry of the Lebesgue integral $\int_{\mathbb R^n} f(y+c) dy =\int_{\mathbb R^n} f(y) dy $. For the second point fair enough, since you were using $\sup$ over $K\subset \mathbb R$ i thought it was obvious that $L^\infty = L^\infty(\mathbb R)$ but I can see that this is not clear. Will edit this in

I finally understood all norms, but how did you do that Chain Rule? I can't reproduce here where the $\sqrt{}$ appears and where $\phi_{1/4}$ appears (I know what $\phi$ is)

@LucasZanella ignoring the constant in the front $C(t) = $something to do with $t$ and $\pi$, we have derivatives of $\phi_{1/4} ( \lambda x)$. Each derivative pulls out one copy of $\lambda$, e.g. $$ \phi_{1/4}(\lambda x) ' = \lambda \phi_{1/4}'(\lambda x) $$ I may have gotten the precise value 1/4 wrong...but its not really important what this number is

How does $\pi$ appears? I think I understood it terribly wrong, because the first derivative with respect to $x$ is $-\frac{2x}{4t}e^{-\frac{x^2}{4t}}$

@LucasZanella I do not recommend actually expanding out what the derivative is, write the function $\phi_t$ in terms of $\phi_{1/4}$ and use chain rule. The constant $C(t)$ is the $1/\sqrt{4\pi t}$ thing times also $\sqrt{\pi}$ to get it to match with $\phi_{1/4}$...so you're right actually there's no $\pi$ but it really doesn't matter

I arrived at $\|\partial^n_x\phi_t\|_{L^2} = \|\partial_x^n \phi_{1/4}\|_{L^2}$ does that mean I can simply say this is less than $C^{n+1}\|\partial_x^n \phi_{1/4}\|_{L^2}$ for some $C^{n+1}$ greater than $1$?

@LucasZanella really? Well the worst it can do is pull out an nth power. I don't think it should exactly cancel for every n but as long as you don't get something that grows faster it's ok yes

Take a look at my update, I don't know what you said about canceling for every $n$ but if I do the rule $\int_{\mathbb R} f(\lambda x)^2 dx = \frac1\lambda \int_{\mathbb R} f^2(y) dy$ I get $\phi_{1/4}$

$\phi_{1/4} = e^{-x^2}/\sqrt{\pi}$ right?

Yes, and you compare with $$\phi_t(x) = \frac1{\sqrt{\pi}}\frac1{\sqrt{4 t}} \exp\left(-(x/\sqrt{4t})^2\right)$$

@LucasZanella your computation in the update is as if you were integrating $$ (\partial_x^n \phi_{1/4} )(\lambda x) $$ which is not the same as integrating $$ (\partial_x^n \phi_{t} )(x) $$

@LucasZanella I should have said, "not the same as integrating $\partial_x^n [\phi_t(x)]$". I think this is clearer. Note $$ \phi_t(x) = \phi_{1/4}(\lambda x)$$ where $\lambda = 1/\sqrt{4t}$. Then chain rule gives $$ \partial_x\phi_t(x)=\partial_x[\phi_{1/4}(\lambda x)] = \lambda [\partial_x\phi_{1/4}](\lambda x)$$ do it $n$ times and then do the $L^2$ norm scaling

I've just arrived now at $(\frac{1}{\sqrt{4\pi t}}\int|\partial_x^n e^{-x^2}|^2)^{1/2}$ which is almost $||\partial_x^n(x)\phi_{1/4}||$, but I don't get how $C^{n+1}$ appears

@LucasZanella, no, that's not right, please see my edit. In words: the rescaling of the derivative is not the derivative of the rescaling. This is the job of chain rule. $\phi_t = \phi_{1/4} \circ L$. $\phi_t' = \phi_{1/4}'\circ L $, times $L'$. Here $L$ is a linear scaling.

@LucasZanella I lied, I thought of a simple example to compute that should convince you. set $f(x) = \exp(-2x)$. $f'(x) = -2\exp(2x)$ and set $g(x) = \exp(-x)$ so that $f = g(2x)$. on $(0,\infty)$, $$\int_0^\infty \exp(-kx) dx = \frac1k$$ just Lebesgue scaling for $n=0$ is ok, $$ \| f\|^2_{L^2} = \int_0^\infty g(2x)^2 dx = \frac14 = \frac12\int_0^\infty g(x)^2 dx$$ chain rule needed for $n=1$. this is like $\|\partial_x \phi_t\|^2_{L^2}:$ $$\|\partial_x f\|_{L^2}^2 = \int_0^\infty |\partial_x f(x)|^2 dx = 2^2 \int_0^\infty f^2 dx = 1 $$

@LucasZanella but if I proceed like your update, (in red is wrong) $$\|\partial_x f\|^2_{L^2} = \int_0^\infty |\partial_x f(x)|^2 dx \color{red}{\overset{?}{=} \int_0^\infty |(\partial_x g)(2x)|^2 dx} = \frac1{2}\int_0^\infty |\partial_x g(x)|^2 dx = \frac12$$

(too late to edit, that last equal sign should be $= 1/4$)