I am studying FEM the very basics. I don't have a very strong background in math nor in functional analysis. Having said that, here's the problem I'm analyzing.
$ -\mu u'' + \sigma u = f ~~~~~ x \in (0,1) $ $u(0) = u(1) = 0$
From that I got the bilinear form of the problem as:
$ a(u, v) = \int_\Omega \mu u'v'dx + \int_\Omega \sigma u vdx $ $ F(v) = \int_\Omega fvdx ~~~~~~~~~~\forall v \in V=H_0^1$
I was trying to understand the proof of continuity of the bilinear form and I found that it could be proven as follows:
$|a(u,v)|= \Big|\int_\Omega \mu u'v'dx + \int_\Omega \sigma u vdx \Big|≤ \Big|\int_\Omega \mu u'v'dx\Big| + \Big|\int_\Omega \sigma u vdx\Big|$ $≤ \mu ||u'||_{L^2} ||v'||_{L^2} + \sigma||u||_{L^2}||v||_{L^2}$ $≤ \max(\mu, \sigma) (||u'||_{L^2} ||v'||_{L^2} + ||u||_{L^2}||v||_{L^2})$
As far as I understand the stuff above is using the Cauchy-Schwarz inequality. Then to complete the verification of the continuity I have to "change" (I don't know if it's the correct word) from $L^2$ to $H^1$ because there's where I defined my space $V$ to be.
So I found two expression that I don't understand how to interpret
$||u||_{L^2} ≤ ||u||_{H^1},~~~~~~~~||u'||_{L^2} ≤ ||u||_{H^1}$
does it mean that a norm in $L^2$ is bounded by a norm in $H^1$ ?? If so, how can I visualize that (geometrically) ?