Overview
I am trying to learn Finite Element Analysis on my own through the course "The Finite Element Method for Problems in Physics" on Coursera. In the course the instructor defined the $H^1$ norm as $$\left\Vert v\right\Vert_1:= \dfrac{1}{m(\Omega)^{\frac{1}{n_{sd}}}}\underset{\Omega}\int{\left((u^h)^2+m(\Omega)^{\frac{2}{n_{sd}}}(u^h_{,x})^2 \mathrm{d}x\right)}$$
$$\begin{align*} \text{where, }&\\ &m(\Omega) &\longrightarrow & \text{ Measure of the domain } \Omega \text{ (in this case the lecturer told that it will be } L^{n_{sd}})\\ &L &\longrightarrow& \text{ Length of the domain for which the Finite Element has been constructed}\\ &n_{sd} &\longrightarrow &\text{ Number of spatial dimensions (consider this to be 1D for now)}\\ \end{align*}$$
After defining the above, the lecturer told that "We have divided through by measure of omega raised to 1 over number of spatial dimensions to get rid of the effect of having integrated, the function and its derivative squared over the domain."
My Question
Why do we divide the whole integration with $m(\Omega)^{\frac{1}{n_{sd}}}$ and why do we multiply $m(\Omega)^{\frac{2}{n_{sd}}}$ with $u^h_x$ in the integrand?
My Attempt at finding solution
As I haven't done a course on functional analysis, so I started searching the web for it. What I found was that the definition of the $H^1{\Omega}$ is as follows: -
$$\left\Vert u \right\Vert^2_{H^1(\Omega)}=\left\Vert u \right\Vert^2_{L^2(\Omega)}+\left\Vert \nabla{u} \right\Vert^2_{L^2(\Omega)}\tag{1}\\$$
Since, $\left\Vert u \right\Vert_{L^p(\Omega)}=\displaystyle\left(\underset{\Omega}\int|v|^p\mathrm{d}x\right)^{\frac{1}{p}}$. Hence, eq. $(1)$ becomes
$$\left\Vert u \right\Vert^2_{H^1(\Omega)}=\left(\underset{\Omega}\int|u|^2\mathrm{d}x\right) + \left(\underset{\Omega}\int\left|\nabla{u}\right|^2\mathrm{d}x\right)\\ \implies \left\Vert u \right\Vert^2_{H^1(\Omega)}=\left(\underset{\Omega}\int|u|^2 + \left|\nabla{u}\right|^2\mathrm{d}x\right)$$$$
As, we have considered the problem to be one-dimensional, hence we have
$$\left\Vert u \right\Vert^2_{H^1(\Omega)}=\underset{\Omega}\int|u|^2 + \left|u_x\right|^2\mathrm{d}x$$
Since, we are dealing with simple 1D basis functions in 1D Finite Element Methods, hence
$$\left\Vert u \right\Vert^2_{H^1(\Omega)}=\underset{\Omega}\int({u}^2 + {u_x}^2)\mathrm{d}x \implies \left\Vert u \right\Vert_{H^1(\Omega)}=\left(\underset{\Omega}\int({u}^2 + {u_x}^2)\mathrm{d}x\right)^{\frac{1}{2}}$$
I reached till this but I am still not clear as to why do we have to do what I have mentioned under the heading My Question?
I am not that thorough with functional analysis so, an intuitive feel for the explanation would be appreciated rather than a rigorous one.
Thanks for reading through my doubt.
This is what I understood and I think the reason is. (Take it with a pinch of salt).
Now, in Finite Element Analysis, while creating the system matrix you want to do the integration on the reference elements instead of the main element. For $1-$D, the reference element is $[0,1]$ and in $2-$D it's the triangle with vertices $(0,0),(1,0)$. and $(0,1)$. The same analogy can be defined for Quadrilateral elements.
Now coming to your question. The first thing I do is, divide my domain into elements. In 1-D it means intervals. Let my element be $[a,b]$. Consider the integral $$ \int_a^bu^2(x)dx. $$ Now, by the change of variables, I move to $[0,1]$. Hence, my transformation is $y=(x-a)/(b-a)$. So, my integral is $$ (b-a)\int_0^1u^2(y)dy. $$ Now, $(b-a)=m(\Omega)$. Similarly, for the derivative I get $$ \frac{1}{(b-a)}\int_0^1u^2_y(y)dy. $$ Now, if you defined the norm as in the Lecture and then do the change of integral, $m(\Omega)$ cancels and you get the norm without weights on the reference element. Hence, I believe the norm defined as this is just for simplicity. As it will help in future computation when you have to calculate the matrices. I believe you will study this concept further in the course.
Now, why is the norm defined this way is equivalent to norm that you found on the internet? There is something called the equivalence of norm and as the norm defined is just a scaled version, the internet norm is equivalent to the lecture norm.