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) ?


1 Answer 1


You need to know the definition of the two norms:

$$ \|u\|_{L^2}^2 = \int_\Omega |u(x)|^2 \,dx $$ 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.

  • $\begingroup$ What do you mean by "adding the missing (non-negative) term"? Where? $\endgroup$
    – BRabbit27
    Nov 28, 2012 at 17:50
  • 1
    $\begingroup$ I wrote down one of the two estimates explicitly; the other follows in exactly the same way. $\endgroup$ Nov 28, 2012 at 17:55

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