# An inequality in the proof of characterization of the $H^{-1}$ norm in Evans's PDE book

I’m reading through the Sobolev Spaces section in Evans’s Partial Differential Equations book, and I was stuck on a theorem characterizing the $H^{-1}$ norm. On page 299 Theorem 1 (in the second edition), he proves that for $f \in H^{-1}(U)$,
$$\begin{multline}\|f\|_{H^{-1}(U)} = \inf \Big\{ \left( \int_U \sum_{i=0}^n |f^i|^2 dx \right): \langle f, v \rangle = \int_U f^0 v + \sum_{i=1}^n f^i v_{x_i} dx,\\ \;f^0,\ldots,f^n \in L^2(U), \; \forall v \in H^1(U)\Big\}. \end{multline}$$

Here $\langle \cdot, \cdot \rangle$ denotes the dual pairing of $H^{-1}$ and $H_0^1$.

To show this, he first states that given $f \in H^{-1}(U)$, we can apply to the Riesz Representation Theorem to get an element $u \in H_0^1$ such that $(u,v)_{H_0^1} = \langle f,v \rangle \; \forall v \in H_0^1$, where $(u,v)_{H_0^1} = \int Du \cdot Dv + uv dx$ is an inner product on $H_0^1$. Then we can define $f^0=u$ and $f^i = u_{x_i}$ for each $i$.

Now, if $\langle f, v \rangle = \int_U g^0 v + \sum_{i=1}^n g^i v_{x_i} dx$, for $g^0,\ldots,g^n \in L^2(U)$, then by choosing $v=u$, we can see that
$$\int_U |Du|^2 +|u|^2 dx \leq \int_U \sum_{i=0}^n |g^i|^2 dx.$$

How does he get this inequality?

When you let $v=u$, you get
$$\int_U |Du|^2 +|u|^2 dx = \int_U g^0 u + \sum_{i=1}^n g^i u_{x_i} dx$$
and I don’t see how you can get the $\{u,u_{x_i}\}$ terms to disappear on the right side.

#### Solutions Collecting From Web of "An inequality in the proof of characterization of the $H^{-1}$ norm in Evans's PDE book"

It’s an application of Cauchy-Schwarz inequality(ies). Since $u$, $u_{x_i}$ and $g^i$are in $L^2(U)$ we have $\int_ug^0udx\leq \sqrt{\int_U|g^0|^2dx}\sqrt{\int_U|u|^2dx}$ and for all $i\in\{1,\ldots, n\}:\int g^iu_{x_i}\leq \sqrt{\int_U|g^i|^2dx}\sqrt{\int_U|u_{x_i}|^2dx}$, hence
\begin{align*}
\lVert u\rVert_{H_0^1(U)}^2&=\int_U g^0udx+\sum_{i=0}^ng^iu_{x_i}dx\\
&\leq \sqrt{\int_U|g^0|^2dx}\sqrt{\int_U|u|^2dx}+\sum_{i=1}^n\sqrt{\int_U|g^i|^2dx}\sqrt{\int_U|u_{x_i}|^2dx} \\
&\leq \sqrt{\int_U|g^0|^2dx+\sum_{i=1}^n\int_U |g^i|^2dx}\sqrt{\int_U|u|^2dx+\sum_{i=1}^n\int_U|u_{x_i}|^2dx},
\end{align*}
applying this time Cauchy-Schwarz inequality for sums, namely $\displaystyle\sum_{i=0}^n a_ib_i\leq\sqrt{\sum_{i=0}^n|a_i|^2}\sqrt{\sum_{i=0}^n|b_i|^2}$, for
$\displaystyle a_0=\sqrt{\int_U|g^0|^2dx}$, $\displaystyle b_0=\sqrt{\int_U|u|^2dx}$, $\displaystyle a_i=\sqrt{\int_U|g^i|^2dx}$ and $\displaystyle b_i=\sqrt{\int_U|u_{x_i}|^2dx}$ for $1\leq i\leq n$.
If $u=0$, the inequality we have to show is obvious, and if it’s not the case we
can divide by $\displaystyle\sqrt{\int_U|u|^2dx+\sum_{i=1}^n\int_U|u_{x_i}|^2dx}$.

As Evans says in the book, setting (3) and using (5), one has
$$\int_U|Du|^2+|u|^2\ dx\leq\int_U |g^0 u|+\sum_{i=1}^n |g^iu_{x_i}|\ dx=:I\tag{1}$$
All you need now is just
$$|ab|\leq\frac{1}{2}(a^2+b^2).$$
Applying this to each term of the integrand on the RHS of (1), we have
$$I\leq \frac{1}{2}\int_U\sum_{i=0}^n|g^i|^2+|u|^2+|Du|^2\ dx\tag{2}$$
Combining (1) and (2), we are done.

(3) and (5) are as the following in the book: