# Norm in L2 bounded by norm in H1

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) ?

$$\|u\|_{L^2}^2 = \int_\Omega |u(x)|^2 \,dx$$ and $$\|u\|_{H^1}^2 = \|u'\|_{L^2}^2 + \|u\|_{L^2}^2.$$ Hence in each of the two expressions, by adding the other term (which is non-negative since it's a norm) you get the upper bound $\|u\|_{H^1}^2$; e.g., $$\|u\|_{L^2}^2 \leq \|u\|_{L^2}^2 + \|u'\|_{L^2}^2 = \|u\|_{H^1}^2.$$ Taking the square root gives you the desired bound.
(Note that $\|u'\|_{L^2}$ is not a norm on $H^1$, since it is zero for any (non-zero) constant $u$.)
I'm not sure that one can visualize the $H^1$ norm geometrically (at least I can't), but you can think of the $L^2$ part of the norm as measuring the magnitude and the $H^1$ part as measuring the oscillation of a function.