# An equality in Evans' PDE book chapter 9

The following are the necessary background definitions and steps in his book(p. 529-530, PDE, second edition).

More precisely, assume that the functions $$w_{k}=w_{k}(x)$$ $$(k=1, \ldots)$$ are smooth and $$\left\{w_{k}\right\}_{k=1}^{\infty} \text { is an orthonormal basis of } H_{0}^{1}(U)$$ taken with the inner product $$(u, v)=\int_{U} D u \cdot D v d x$$. (We could for instance take $$\left\{w_{k}\right\}_{k=1}^{\infty}$$ to be the set of appropriately normalized eigenfunctions for $$-\Delta$$ in $$\left.H_{0}^{1}(U) \cdot\right)$$ We will look for a function $$u_{m} \in H_{0}^{1}(U)$$ of the form $$u_{m}=\sum_{k=1}^{m} d_{m}^{k} w_{k}\tag{6}$$ where we hope to select the coefficients $$d_{m}^{k}$$ so that $$\int_{U} \mathbf{a}\left(D u_{m}\right) \cdot D w_{k} d x=\int_{U} f w_{k} d x \quad(k=1, \ldots, m)\tag{7}$$ THEOREM 1 (Construction of approximate solutions). For each integer $$m=1, \ldots$$, there exists a function $$u_{m}$$ of the form (6) satisfying the identities (7).

Proof. Define the continuous function $$\mathbf{v}: \mathbb{R}^{m} \rightarrow \mathbb{R}^{m}, \mathbf{v}=\left(v^{1}, \ldots, v^{m}\right)$$, by setting $$v^{k}(d):=\int_{U} \mathbf{a}\left(\sum_{j=1}^{m} d_{j} D w_{j}\right) \cdot D w_{k}-f w_{k} d x \quad(k=1, \ldots, m)\tag{10}$$ for each point $$d=\left(d_{1}, \ldots, d_{m}\right) \in \mathbb{R}^{m}$$. Now \begin{aligned} \mathbf{v}(d) \cdot d &=\int_{U} \mathbf{a}\left(\sum_{j=1}^{m} d_{j} D w_{j}\right) \cdot\left(\sum_{j=1}^{m} d_{j} D w_{j}\right)-f\left(\sum_{j=1}^{m} d_{j} w_{j}\right) d x \\ & \geq \int_{U} \alpha\left|\sum_{j=1}^{m} d_{j} D w_{j}\right|^{2}-\beta-f\left(\sum_{j=1}^{m} d_{j} w_{j}\right) d x \quad \text { by }(5) \\ &=\alpha|d|^{2}-\beta|U|-\sum_{j=1}^{m} d_{j} \int_{U} f w_{j} d x \end{aligned} Transcribed from Screenshot1 and Screenshot2

My question is, why the last equality can hold. To be more specific, why do we have that $$\int_U \vert \sum_{j=1}^m d_j Dw_j \vert^2 dx=\vert d \vert^2$$, where $$Dw_j=(\partial_{x_1} w_j, ... , \partial_{x_n} w_j).$$

My attempt: $$\vert \sum_{j=1}^m d_j Dw_j \vert^2=\vert (\sum_{j=1}^m d_j \partial_{x_1} w_j, ... , \sum_{j=1}^m d_j \partial_{x_n} w_j) \vert^2 = (\sum_{j=1}^m d_j \partial_{x_1} w_j)^2+ ... + (\sum_{j=1}^m d_j \partial_{x_n} w_j)^2.$$

Up to this point, I don't know how to proceed to prove that it is greater than or equal to $$\vert d \vert ^2$$. Any help will be appreciated.

The result follows from $$\{w_j\}$$ being an orthonormal set of vectors in $$H^1_0(U)$$ with inner product given by $$(u,v)=\int_U Du\cdot Dv dx.$$ Indeed, \begin{align*} \int_U \bigg \vert \sum_{i=1}^m d_i Dw_i \bigg \vert^2 d x &= \int_U \bigg ( \sum_{i=1}^m d_i Dw_i \bigg ) \cdot \bigg ( \sum_{j=1}^m d_j Dw_j \bigg ) d x \\ &=\sum_{i,j=1}^m d_i d_j \int_U D w_i \cdot D w_j d x \\ &= \sum_{i,j=1}^m d_i d_j \delta_{ij} \\ &= \vert d \vert^2 \end{align*} where $$\delta_{ij}$$ is the Kronecker delta.
• Thanks for the quick and neat answer. I understand your argument. But Evans didn’t explicitly define the inner product on $H_0^1$ and he only defined the inner product on $H^1$. In his definition there is an additional term $\int fg dx$. And these two inner products can not dominate each other. If the inner product is defined in Evans’ way, can we find an alternative way to prove this equality? Thanks. Sep 5, 2021 at 13:16
• In the screenshot you've attached the first sentence says that the inner product he is taking on $H^1_0(U)$ is the one I wrote in my solution. On $H^1_0$ this inner product is equivalent to the standard one due to the Poincare inequality. Sep 5, 2021 at 13:22