# Taylor

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bio website mathtm.blogspot.com location University of California Los Angeles, CA age member for 1 year, 10 months seen Mar 2 at 20:54 profile views 215

I am a recent graduate in applied mathematics at UCLA and currently trying to break into the quantitative investment/trading industry while also continuing to pursue advanced graduate-level mathematics as both a hobby and career necessity.

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 Mar23 comment Norm of the linear functional Estimating $||A||$ by the triangle inequality (as the OP did) also shows $||A||\leq||\phi||_{1}$, so in fact $||A||=||\phi||_{1}$. This applies to convolutions as well, a common class of integral operators, so is helpful fact to keep in mind, since information about $A$ can be obtained from information about its kernel $\phi$. Mar23 comment Norm of the linear functional I like your explicit construction. Alternatively though, if the OP has some experience with measure theory, one can simply use the fact that every measurable function is an a.e. p.w. limit of continuous functions $\phi_{n}$. Since the sign function of $\phi$ is measurable, the result follows from dominated convergence: $$||A||\geq\lim_{n\to\infty}\int_{a}^{b}\phi_{n}\phi\;dx=\int_{a}^{b}|\phi(x)\|;‌​dx=||\phi||_{1}.$$ This technique comes up in a lot of situations involving integral operators and uniform boundedness principle; the most famous is Fourier series and the Dirichlet kernel. Mar21 comment Schwarz Reflection Prinnciple for (Real) Harmonic Functions I understand your point, and thanks for posting your answer, especially since it led to the citation of a free online text; I had no idea Sheldon Axler had other text books besides his linear algebra one, much less on harmonic functions! With respect to this problem, I think anyone that has read through the responses/question will easily be able to finish the proof with a continuity argument. In either case, for purposes of this problem, it is solved by part (b) using the Poisson kernel anyhow. Mar13 accepted Necessary Condition for $C^{2}$ Regularity of this Function Mar13 asked Learning Aid for Basic Theorems of Topological Vector Spaces in Functional Analysis Mar6 asked Necessary Condition for $C^{2}$ Regularity of this Function Mar5 awarded Benefactor Mar5 accepted Uniqueness of PDE BVP/IVP Modified Wave Equation Mar5 revised Uniqueness of PDE BVP/IVP Modified Wave Equation added 814 characters in body Mar5 comment Uniqueness of PDE BVP/IVP Modified Wave Equation I posted a correct (I hope) solution with the estimate. I was applying Gronwall's inequality to an estimate of the form $E_{2}'\leq \phi E_{2}+\psi$ intead of $E_{2}'\leq\phi E_{2}.$ In the former, you get an extra term which is hard to work with (in particular, to show it vanishes), but in the ladder you only get a product with one factor being $E_{2}(0)$ which of course vanishes due to the boundary conditions on $u:=u_{1}-u_{2}.$ Mar5 revised Uniqueness of PDE BVP/IVP Modified Wave Equation added 814 characters in body Mar5 revised Uniqueness of PDE BVP/IVP Modified Wave Equation added 1505 characters in body Mar5 comment Uniqueness of PDE BVP/IVP Modified Wave Equation Nevermind, you're right -- you can't subtract them. I would need a lower bound on one and an upper bound on the other. Mar5 revised Uniqueness of PDE BVP/IVP Modified Wave Equation deleted 2622 characters in body Mar5 revised Uniqueness of PDE BVP/IVP Modified Wave Equation added 36 characters in body Mar5 comment Uniqueness of PDE BVP/IVP Modified Wave Equation (See original post under the edit for my response -- it was too much to put into a comment(s)). Mar5 revised Uniqueness of PDE BVP/IVP Modified Wave Equation added 1456 characters in body Mar5 revised Uniqueness of PDE BVP/IVP Modified Wave Equation added 1456 characters in body Mar4 comment Uniqueness of PDE BVP/IVP Modified Wave Equation With that energy you get $E'(t)=4\int_{U}uu_{t}q\;dx$ instead of $E'(t)=0$! Mar4 comment Uniqueness of PDE BVP/IVP Modified Wave Equation I'll toy around a little more and see if I can't get $E$ and $E'$ to both appear in the same expression in order to apply Gronwall's inequality.