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Let $G:=\{(x_1,x_2)\in\mathbb{R}^2:0\leq x_1\leq 1\}$ be a streak in $\mathbb{R}^2$ and for $f\in \mathcal{C}^1(G,\mathbb{C})$ we define

$$\|f\|_{1,\infty}:=\sup_{x\in G}|f(x)|+\sup_{x\in G}|\nabla f(x)|,\qquad \|f\|_\sim:=\sup_{x\in G}|\nabla f(x)|.$$

Further, the linear space $E$ is defined by

$$E:=\{f\in\mathcal{C}^1(G,\mathbb{C})\ \big|\ \|f\|_{1,\infty}<\infty \land \forall x_2\in\mathbb{R}:f(0,x_2)=0=f(1,x_2)\}.$$

I should show that $\|\cdot\|_\sim$ and $\|\cdot\|_{1,\infty}$ are equivalent.

I have to show that there exist $c_1,c_2>0$ so that for all $f\in E$

  • $\|f\|_\sim \leq c_1\cdot\|f\|_{1,\infty}$ and
  • $\|f\|_{1,\infty}\leq c_2\cdot\|f\|_\sim$.

I've started to show the last one, but I'm not sure with my proof:

\begin{align*} \|f\|_{1,\infty} &= \sup_{x\in G}|f(x)|+\sup_{x\in G}|\nabla f(x)|\\ &=\sup_{x\in G}|f(x)|+\|f\|_{\sim}\\ &\leq\sup_{x\in G}|f(x)|\cdot\|f\|_\sim+\|f\|_\sim\\ &=\|f\|_\sim\cdot\left(\sup_{x\in G}|f(x)|+1 \right)=: c_2\cdot \|f\|_\sim \end{align*}

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It is clear that $\|f\|_\sim \leq \|f\|_{1,\infty}$ by the very definitions of both norms. So we show the second inequality

A priori, $||f||_\sim$ could be smaller that $1$, so that your proof is wrong. Instead, if $f \in C^1(G,\mathbb{R})$, and let $x=(x_1,x_2) \in G$. Therefore $f(0,x_2) = 0$ by the definition of $G$, and using the mean value theorem as $f$ is continuously differentiable, we have the following: \begin{align*} |f(x)|&=|f(x) - f(0,x_2)| \leq |(x_1,x_2) - (0,x_2)|\cdot \sup_{x\in [(x_1,x_2), (0,x_2)]}|\nabla f| \leq \sup_{x\in G}|\nabla f| \end{align*}

where $[(x_1,x_2), (0, x_2)]$ denotes the segment between the two points. Hence, we have that $\|f\|_{1,\infty}\leq 2\cdot\|f\|_\sim$.

If $f \in C^1(G,\mathbb{C})$, consider $f_1 = \mathfrak{Re}(f)$ and $f_2 = \mathfrak{Im}(f)$, and $\mathbb{R}^2$ equipped with $||\cdot||_\infty$ identified with the complex plane. Then \begin{align*} |f(x)| &\leq \sup_{x\in G} |\nabla f_1(x)| + \sup_{x\in G} |\nabla f_2(x)|\\ &\leq 2\cdot \max\{\sup_{x\in G}|\nabla f_1 (x)|,\sup_{x\in G}|\nabla f_2 (x)|\}\\ &\leq 2\cdot \sup_{x\in G}\max\{|\nabla f_1(x)|,|\nabla f_2(x)|\}\\ &\leq 2\cdot \sup_{x\in G}\max\{\sup_{y\in \partial D(0,1)}|\nabla f_1(x)\cdot y|,\sup_{y\in \partial D(0,1)}|\nabla f_2(x)\cdot y|\}\\ &\leq 2 \cdot \sup_{x\in G}\sup_{y\in \partial D(0,1)}\max\{|\nabla f_1(x)\cdot y|,|\nabla f_2(x)\cdot y|\}\\ &\leq 2 \cdot \sup_{x\in G}|\nabla f(x)| \end{align*}

Therefore, define $c_2 := 3$.

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  • $\begingroup$ I have a question: Why do you use the mean value theorem? I know that this theorem is only true for real valued functions, but I have here a complex valued function. $\endgroup$ – MathCracky Oct 23 '16 at 16:21
  • $\begingroup$ @MathCracky Fixed. $\endgroup$ – Hermès Oct 23 '16 at 17:36

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