# proof of weak maximum principle for real domains

First, we show that, if $\mathrm{\Delta}f>0$ (where $\mathrm{\Delta}$ denotes the Laplacian^{} on ${\mathbb{R}}^{d}$) on $K$, then $f$ cannot attain a maximum on the interior of $K$. Assume, to the contrary, that $f$ did attain a maximum at a point $p$ located on the interior of $K$. By the second derivative test^{}, the matrix of second partial derivatives^{} of $f$ at $p$ would have to be negative semi-definite. This would imply that the trace of the matrix is negative. But the trace of this matrix is the Laplacian, which was assumed to be strictly positive on $K$, so it is impossible for $f$ to attain a maximum on the interior of $K$.

Next, suppose that $\mathrm{\Delta}f=0$ on $K$ but that $f$ does not attain its maximum on the boundary of $K$. Since $K$ is compact, $f$ must attain its maximum somewhere, and hence there exists a point $p$ located in the interior of $K$ at which $f$ does attain its maximum. Since $K$ is compact, the boundary of $K$ is also compact, and hence the image of the boundary of $K$ under $f$ is also compact. Since every element of this image is strictly smaller than $f(p)$, there must exist a constant $C$ such that $$ whenever $x$ lies on the boundary of $K$. Furthermore Since $K$ is a compact subset of ${\mathbb{R}}^{d}$, it is bounded. Hence, there exists a constant $R>0$ so that $$ for all $x\in K$.

Consider the function^{} $g$ defined as

$$g(x)=f(x)+(f(p)-C)\frac{{|x-p|}^{2}}{{R}^{2}}$$ |

At any point $x\in K$,

$$ |

In particular, if $x$ lies on the boundary of $K$, this implies that

$$ |

Since $g(p)=f(p)$ this inequality^{} implies that $g$ cannot attain a maximum on the boundary of $K$.

This leads to a contradiction. Note that, since $\mathrm{\Delta}f=0$ on $K$,

$$\mathrm{\Delta}g=\frac{d(f(p)-C)}{{R}^{2}}>0$$ |

which implies that $g$ cannot attain a maximum on the interior of $K$. However, since $K$ is compact, $g$ must attain a maximum somewhere on $K$. Since we have ruled out both the possibility that this maximum occurs in the interior and the possibility that it occurs on the boundary, we have a contradiction. The only way out of this contradiction is to conclude that $f$ does attain its maximum on the boundary of $K$.

Title | proof of weak maximum principle for real domains |
---|---|

Canonical name | ProofOfWeakMaximumPrincipleForRealDomains |

Date of creation | 2013-03-22 14:35:21 |

Last modified on | 2013-03-22 14:35:21 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 6 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 30F15 |

Classification | msc 31B05 |

Classification | msc 31A05 |

Classification | msc 30C80 |