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ifw
ifw
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One cannot bound $\sup u$ in terms of $c$. To see this, set $u(x) = c + \alpha \,\varphi(\varepsilon x)$, where $\varphi\in\mathcal S(\mathbb R^2;\mathbb R)$$\varphi\in S(\mathbb R^2;\mathbb R)$, $\varphi(0)=1$, and $\varphi\leq1$, while $\alpha>0$, $\varepsilon>0$ are parameters. Notice that $$ \Delta u(x) + e^{-u(x)} \geq e^{-\alpha} \left( e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c} \right). $$

Now, given $K>c_+$, let $\alpha=K-c>0$ and choose $\varepsilon>0$ in such a way that $e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c}\geq0$. Then $\sup u= u(0)=K$ and $\Delta u + e^{-u}\geq0$; so we have the desired example.

One cannot bound $\sup u$ in terms of $c$. To see this, set $u(x) = c + \alpha \,\varphi(\varepsilon x)$, where $\varphi\in\mathcal S(\mathbb R^2;\mathbb R)$, $\varphi(0)=1$, and $\varphi\leq1$, while $\alpha>0$, $\varepsilon>0$ are parameters. Notice that $$ \Delta u(x) + e^{-u(x)} \geq e^{-\alpha} \left( e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c} \right). $$

Now, given $K>c_+$, let $\alpha=K-c>0$ and choose $\varepsilon>0$ in such a way that $e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c}\geq0$. Then $\sup u= u(0)=K$ and $\Delta u + e^{-u}\geq0$; so we have the desired example.

One cannot bound $\sup u$ in terms of $c$. To see this, set $u(x) = c + \alpha \,\varphi(\varepsilon x)$, where $\varphi\in S(\mathbb R^2;\mathbb R)$, $\varphi(0)=1$, and $\varphi\leq1$, while $\alpha>0$, $\varepsilon>0$ are parameters. Notice that $$ \Delta u(x) + e^{-u(x)} \geq e^{-\alpha} \left( e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c} \right). $$

Now, given $K>c_+$, let $\alpha=K-c>0$ and choose $\varepsilon>0$ in such a way that $e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c}\geq0$. Then $\sup u= u(0)=K$ and $\Delta u + e^{-u}\geq0$; so we have the desired example.

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ifw
ifw
  • 1.2k
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  • 18

One cannot bound $\sup u$ in terms of $c$. To see this, set $u(x) = c + \alpha \,\varphi(\varepsilon x)$, where $\varphi\in\mathcal S(\mathbb R^2;\mathbb R)$, $\varphi(0)=1$, and $\varphi\leq1$, while $\alpha>0$, $\varepsilon>0$ are parameters. Notice that $$ \Delta u(x) + e^{-u(x)} \geq e^{-\alpha} \left( e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c} \right). $$

Now, given $K>c_+$, let $\alpha=K-c>0$ and choose $\varepsilon>0$ in such a way that $e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c}\geq0$. Then $\sup u\geq u(0)=K$$\sup u= u(0)=K$ and $\Delta u + e^{-u}\geq0$; so we have the desired example.

One cannot bound $\sup u$ in terms of $c$. To see this, set $u(x) = c + \alpha \,\varphi(\varepsilon x)$, where $\varphi\in\mathcal S(\mathbb R^2;\mathbb R)$, $\varphi(0)=1$, and $\varphi\leq1$, while $\alpha>0$, $\varepsilon>0$ are parameters. Notice that $$ \Delta u(x) + e^{-u(x)} \geq e^{-\alpha} \left( e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c} \right). $$

Now, given $K>c_+$, let $\alpha=K-c>0$ and choose $\varepsilon>0$ in such a way that $e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c}\geq0$. Then $\sup u\geq u(0)=K$ and $\Delta u + e^{-u}\geq0$; so we have the desired example.

One cannot bound $\sup u$ in terms of $c$. To see this, set $u(x) = c + \alpha \,\varphi(\varepsilon x)$, where $\varphi\in\mathcal S(\mathbb R^2;\mathbb R)$, $\varphi(0)=1$, and $\varphi\leq1$, while $\alpha>0$, $\varepsilon>0$ are parameters. Notice that $$ \Delta u(x) + e^{-u(x)} \geq e^{-\alpha} \left( e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c} \right). $$

Now, given $K>c_+$, let $\alpha=K-c>0$ and choose $\varepsilon>0$ in such a way that $e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c}\geq0$. Then $\sup u= u(0)=K$ and $\Delta u + e^{-u}\geq0$; so we have the desired example.

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ifw
ifw
  • 1.2k
  • 1
  • 6
  • 18

One cannot bound $\sup u$ in terms of $c$. To see this, set $u(x) = c + \alpha \,\varphi(\varepsilon x)$, where $\varphi\in\mathcal S(\mathbb R^2;\mathbb R)$, $\varphi(0)=1$, and $\varphi\leq1$, while $\alpha>0$, $\varepsilon>0$ are parameters. Notice that $$ \Delta u(x) + e^{-u(x)} \geq e^{-\alpha} \left( e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c} \right). $$

Now, given $K>c_+$, let $\alpha=K-c>0$ and choose $\varepsilon>0$ in such a way that $e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c}\geq0$. Then $\sup u\geq u(0)=K$ and $\Delta u + e^{-u}\geq0$; so we have the desired example.