Show that bilinear form A(u , v) is continuous

Question

I want to show that the following bilinear form is continuous:

$A:H^{1}(0,1)\times H^{1}(0,1)\rightarrow \mathbb{R}$, where $H^{1}(0,1)$ is the Sobolev space and $A$ is define by

$A(u , v) = \int_{0}^{1}\alpha(x)u'(x)v'(x)dx + \int_{0}^{1}u(x)v(x)dx + k_{0}u(1)v(1)$, where $k_{0}>0$ and $\alpha$ is bounded and positive.

That's what I did:

$|A(u , v)|=|\int_{0}^{1}\alpha(x)u'(x)v'(x)dx + \int_{0}^{1}u(x)v(x)dx + k_{0}u(1)v(1)|\leq |\int_{0}^{1}\alpha(x)u'(x)v'(x)dx| + |\int_{0}^{1}u(x)v(x)dx| + |k_{0}u(1)v(1)|$

I know that we can use the vectors of $\mathbb{R}^{2}$ $x_{1}=(||u||_{L_{2}},||u'||_{L_{2}})$ and $x_{2}=(||v||_{L_{2}},||v'||_{L_{2}})$ to conclude the proof when there is no constant $|k_{0}u(1)v(1)|$, but I don't know how to finish the proof from here with that constant.

Any tips?

Thanks a lot

Answer 1

First you should ask yourself why the bilinear form is even well-defined. Functions in Sobolev spaces are generally not continuous and you can't talk about their values at some point. However, in the one-dimensional case, functions in $H^1(0,1)$ turn out to be (absolutely) continuous and you have an estimate of the form $||f||_{\infty} \leq C ||f||_{H^1}$ (so the inclusion $H^1(0,1) \rightarrow C^0([0,1])$ is continuous). This is the content of the one-dimensional version of the (one of the various) Sobolev embedding theorem.

Using such an estimate, you can bound $|k_0u(1)v(1)|$ by $k_0 C^2 ||u||_{H^1} ||v||_{H^1}$.