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2d
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revised |
Rudin Theorem 1.17 [Edit removed during grace period] |
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2d
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answered | Rudin Theorem 1.17 |
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May
13 |
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comment |
Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$ This integral has no closed form (elementary) anti-derivative, if that's what you mean by closed form solution. Certainly without that, you cannot evaluate the integral via FTC. A simple calculation shows that your integral is equivalent to $$\int_{0}^{\frac{\pi}{2}}\log\Bigg(\log\Bigg(\cot\Bigg(\frac{t}{2}\Bigg)\Bigg) \Bigg) \cos(x)\;dx$$ if you aren't convinced an elementary anti-derivative does not exist. |
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May
9 |
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awarded | Yearling |
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May
9 |
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awarded | Yearling |
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Apr
12 |
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asked | Taking Graduate Quals as an Undergrad to prepare for Grad School Applications |
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Mar
24 |
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comment |
Norm of the linear functional @Julien, see my previous comment about the harder direction. And I often see the estimate $\int f\leq\int|f|$ termed "triangle inequality," which is exactly what the OP used to obtain the easy estimate. |
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Mar
23 |
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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 $\phi$. |
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Mar
23 |
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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. |
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Mar
21 |
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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. |
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Mar
14 |
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awarded | Student |
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Mar
14 |
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asked | How to Analyze Applied Forces/Torques of System With Multiple Massless/Frictionless Pulleys of Different Radii |
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Mar
13 |
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accepted | Necessary Condition for $C^{2}$ Regularity of this Function |
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Mar
13 |
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asked | Learning Aid for Basic Theorems of Topological Vector Spaces in Functional Analysis |
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Mar
6 |
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asked | Necessary Condition for $C^{2}$ Regularity of this Function |
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Mar
5 |
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awarded | Benefactor |
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Mar
5 |
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accepted | Uniqueness of PDE BVP/IVP Modified Wave Equation |
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Mar
5 |
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revised |
Uniqueness of PDE BVP/IVP Modified Wave Equation added 814 characters in body |
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Mar
5 |
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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 $u:=u_{1}-u_{2}.$ |
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Mar
5 |
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revised |
Uniqueness of PDE BVP/IVP Modified Wave Equation added 814 characters in body |
