A Proof of Liouville’s Theorem in complex analysisAugust 31, 2009

Posted by Phi. Isett in Uncategorized.
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Liouville’s theorem in complex analysis says that the only bounded holomorphic functions are constant.  This is my favorite

Proof:

Pretend you had a non-constant, bounded holomorphic function $f$ on all of $\mathbb{C}$.  Since $f$ is bounded at $\infty$, Riemann’s theorem on removable singularities implies that $f$ extends to a holomorphic (and hence continuous) function on the Riemann sphere $\mathbb{C} \cup \{ \infty \}$, which is compact.  If $f$ were not constant, the open mapping theorem would apply to this extension, and the image of the Riemann sphere would be an open subset of $\mathbb{C}$.  But this cannot be the case, because $\mathbb{C}$ has no nonempty subsets which are both open and compact.  $\Box$

There are some downsides to this proof.  It does not rely on the theory of Riemann surfaces (not a downside).  It does, however, rely on some relatively (though not truly) heavy machinery, and in order to be a correct proof, one isn’t allowed to use Liouville’s theorem to develop this machinery (but it is possible).  Liouville’s theorem follows rather immediately from Cauchy’s integral formulae, and I don’t personally know how to establish things like analyticity, Riemann’s theorem and the open mapping theorem without this tool (although I would be interested if anyone else does!).  I also don’t think it really generalizes very well to other PDE for which a Liouville theorem holds.

On the geometric meaning of the Cauchy Schwarz inequality, an intro to exterior powers, and surface integralsAugust 19, 2009

Posted by Phi. Isett in Uncategorized.
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This is a brief remark on the Cauchy Schwarz inequality and one way of understanding its geometric meaning (at least in the context of a real inner product space).  In a real inner product space $V$, the inner product $<\cdot, \cdot>$ allows for a generalization of intuitive geometric notions of “length”, “angle”, and “perpendicular” for vectors in $V$.  For two elements $u, v \in V$, I will write $u \wedge v$ to indicate the parallelogram formed by taking $u$ and $v$ as edges.  The reader may as well simply imagine $V = {\mathbb{R}}^n$ for his favorite $n$ and that the inner product is the usual dot product.
The Cauchy Schwarz inequality says that the area of a parallelogram $u \wedge v$ is positive unless u and v are co-linear (it is also equivalent to the triangle inequality, but I will be talking about this formulation instead).  If $u$ and $v$ are co-linear (they point in the same direction or perhaps opposite directions; maybe $v = - u$) the parallelogram one forms with these two vectors is degenerate and has zero area; otherwise, you wind up with a parallelogram which has positive area.  In fact, the volume of $u \wedge v$ is $|| u || \cdot || v ||$ when the two vectors $u$ and $v$ are perpendicular.  If $u$ and $v$ fail to be perpendicular, then we can observe that shifting the edge $u$ by any amount in the direction of the other edge $v$ does not change the area of the parallelogram; thus the area of $A(u \wedge v)$ is the same as the area $A [ (u + \alpha v) \wedge v ]$ for any $\alpha \in \mathbb{R}$.  By choosing $\alpha$ to minimize the length of the first edge (pick $\alpha = - \frac{}{\|v\|^2}$), we can make both edges perpendicular.  The area of the resulting parallelogram must be non-negative (and must be positive when $u$ and $v$ are linearly independent), giving the inequality $\| (u + \alpha v) \|^2 \| v \|^2 = \|u\|^2\|v\|^2 - | |^2 \geq 0$.  (This is actually a self-contained proof of the Cauchy-Schwartz inequality, and it’s the usual proof, but with some motivation regarding how to choose $\alpha$.)
Of course, the Cauchy-Schwartz inequality is also equivalent to the triangle inequality, and the relationship between these two geometric interpretations can be seen by inspecting the parallelogram one forms by drawing “$u + v = v + u$“.
I hope to write on how formalizing this area-of-parallelogram concept works (only on an intuitive level, for now).  For those already familiar with exterior powers of vector spaces and exterior algebras (and their geometric meaning in terms of parallelograms), I can cut to the chase and say that there’s an induced inner product on the exterior powers and the Cauchy Schwartz inequality is what you get by writing $< u \wedge v , u \wedge v > \geq 0$.