20260226_NumPDE

Show that $l_{4}$ is continuous on $H^{1} (Ω)$ , where $l_{4} (v) := \int_{Ω} v (\frac{x}{∥ x ∥}) d x, Ω := {| | x | | \leq 1}$

To show that $l_{4}$ is continuous, we need to show that

| l_{4} (v) | \leq C_{Ω} ∥ v ∥_{H^{1} (Ω)}

First, we simplify $l_{4}$ :

\begin{aligned} l_{4} & = \int_{Ω} v (\frac{x}{∥ x ∥}) d x \\ = \int_{0}^{2 π} \int_{0}^{1} r \cdot v (\frac{r [\begin{array}{c} \cos ϕ \\ \sin ϕ \end{array}]}{r}) d r d ϕ \\ = \int_{0}^{2 π} {[\frac{r^{2}}{2}]}_{0}^{1} \cdot v ([\begin{array}{c} \cos ϕ \\ \sin ϕ \end{array}]) d ϕ \\ = \frac{1}{2} \int_{\partial Ω} v (x) d x \end{aligned}

Thus,

\begin{aligned} | l_{4} (v) | & = \frac{1}{2} \cdot | \int_{\partial Ω} 1 \cdot v (x) d x \\ \leq \frac{1}{2} \cdot {(\int_{\partial Ω} | 1 |^{2})}^{1 / 2} {(\int_{\partial Ω} | v (x) |^{2})}^{1 / 2} & cauchy-schwarz \\ = \frac{1}{2} \cdot \sqrt{| \partial Ω |} \cdot ∥ v (x) ∥_{L^{2} (\partial Ω)} & def. ∥ \cdot ∥_{L^{2}} \\ \leq \frac{\sqrt{| \partial Ω |}}{2} \cdot \sqrt{C} \sqrt{∥ v (x) ∥_{L^{2} (Ω)}} \sqrt{∥ v (x) ∥_{H^{1} (Ω)}} & multipl. trace ineq. \\ \leq \frac{\sqrt{| \partial Ω |}}{2} \cdot \sqrt{C} ∥ v (x) ∥_{H^{1} (Ω)} & ∥ v ∥_{L^{2} (Ω)} \leq ∥ v ∥_{H^{1} (Ω)} \end{aligned}

thus, $l_{4}$ is continuous with $C_{Ω} = \frac{\sqrt{| \partial Ω |}}{2} \sqrt{C}$ .