Activity for Snoopy
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| Edit | Post #290642 |
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| Edit | Post #290643 |
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| Edit | Post #290643 |
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| Edit | Post #290643 |
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| Edit | Post #290643 |
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| Edit | Post #290643 |
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| Edit | Post #290642 |
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| Edit | Post #290643 | Initial revision | — | about 1 year ago |
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A: Fourier transform of an $L^1$ function is uniformly continuous Following the definition of uniform continuity, one basically needs to estimate: $$ \begin{align} |\hat{f}(x)-\hat{f}(y)| &= |\int{\Rbb^n}f(t)(e^{-2\pi i x\cdot t}-e^{-2\pi i y\cdot t})\ dt|\\\\ &\le \intw |f(t)||e^{-2\pi ih\cdot t}-1|\ dt, \end{align} $$ where $h=x-y$. If one can show t... (more) |
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| Edit | Post #290642 | Initial revision | — | about 1 year ago |
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Fourier transform of an $L^1$ function is uniformly continuous $\def\Rbb{\mathbf{R}}$$\def\Cbb{\mathbf{C}}$$\def\intw{\int{\Rbb^n}}$If $f\in L^1(\Rbb^n)$, denote the Fourier transform of $f$ as $$ \hat{f}(x) = \int{\Rbb^n}f(t)e^{-2\pi x\cdot t}\ dt $$ > Problem. If $f\in L^1(\Rbb^n)$, show that $\hat{f}:\Rbb^n\to\Cbb$ is uniformly continuous. Note. Thi... (more) |
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| Edit | Post #290600 |
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| Edit | Post #290599 |
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| Edit | Post #290600 |
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| Edit | Post #290600 | Initial revision | — | about 1 year ago |
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A: For any real number $m$, $ \left|\sum_{n=1}^{\infty}\frac{m}{n^2+m^2}\right|<\frac{\pi}{2} $ Without loss of generality, one can assume that $m>0$. For each positive integer $n$, since the function $x\mapsto\frac{m}{x^2+m^2}$ is decreasing, one has $$ \frac{m}{n^2+m^2} = \int{n-1}^{n}\frac{m}{n^2+m^2}\ dx <\int{n-1}^{n}\frac{m}{x^2+m^2}\ dx $$ It follows that $$ \begin{align} \sum{n=... (more) |
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| Edit | Post #290599 | Initial revision | — | about 1 year ago |
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For any real number $m$, $ \left|\sum_{n=1}^{\infty}\frac{m}{n^2+m^2}\right|<\frac{\pi}{2} $ > Problem. Prove that for any real number $m$, $$ \left|\sum{n=1}^{\infty}\frac{m}{n^2+m^2}\right|<\frac{\pi}{2} $$ Notes. This is an exercise in calculus. There are different ways to approach this problem. One may find the sum on the left explicitly with hyperbolic functions. One can also s... (more) |
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| Edit | Post #290568 |
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| Edit | Post #290583 |
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| Edit | Post #290583 | Initial revision | — | about 1 year ago |
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A: The derivatives of a function at a boundary point Your question is better phrased and answered in Whitney's paper: Analytic extensions of differentiable functions defined in closed sets. Here is an excerpt from the introduction section of the paper, which can be seen from the given AMS link: >Imagealttext (more) |
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| Comment | Post #290576 |
https://en.wikipedia.org/wiki/Whitney_extension_theorem (more) |
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| Edit | Post #290571 |
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| Edit | Post #290571 |
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| Edit | Post #290571 | Initial revision | — | about 1 year ago |
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A: Using convexity in the proof of Hölder’s inequality The first question can be rephrased in a clear way as follows: > Let $A$ and $B$ be two positive real numbers. Show that the function $F(t):=A^{p(1-t)}B^{qt}$ for $t\in[0,1]$ is convex using the convexity of the exponential function. If one rewrites: $$ A^{p(1-t)}B^{qt} = e^{p(1-t)\log(A)+(q... (more) |
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| Edit | Post #290570 | Initial revision | — | about 1 year ago |
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Using convexity in the proof of Hölder’s inequality A key fact for the algebra properties of $L^p$ spaces is Hölder’s inequality: > Let $f \in L^p$ and $g \in L^q$ for some $0 Our task is now to show that $$\intX |fg|\ d\mu \leq 1. \tag{1}$$ >Here, we use the convexity of the exponential function $t \mapsto e^t$ on ${}[0,+\infty)$, which implies ... (more) |
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| Edit | Post #290568 |
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| Edit | Post #290568 |
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| Edit | Post #290568 |
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| Edit | Post #290568 | Initial revision | — | about 1 year ago |
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A: If both $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}f'(x)$ exist, then $\lim_{x\to\infty}f'(x)=0$. Idea. If $\displaystyle \lim{x\to\infty}f'(x)\ne 0$, then for large enough $x$, $f'(x)$ is bounded away from $0$, which implies by the mean value theorem that $|f(x+1)-f(x)|$ is bounded away from $0$. But $|f(x+1)-f(x)|$ must also be small for large $x$ since $\displaystyle \lim{x\to\infty}f(x)$ ex... (more) |
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| Edit | Post #290567 |
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| Edit | Post #290567 | Initial revision | — | about 1 year ago |
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If both $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}f'(x)$ exist, then $\lim_{x\to\infty}f'(x)=0$. Question. Let $f:\mathbf{R}\to\mathbf{R}$ be a differentiable function. If both the limits $\displaystyle \lim{x\to\infty}f(x)$ and $\displaystyle \lim{x\to\infty}f'(x)$ exist, how does one show that $\displaystyle \lim{x\to\infty}f'(x)=0$? Note. This statement is intuitively obvious: if $f$ has a... (more) |
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| Edit | Post #290538 |
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| Edit | Post #290538 |
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| Edit | Post #290538 | Initial revision | — | about 1 year ago |
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A: What does it mean by saying that $C([0,1])$ is a subset of $L^\infty([0,1])$? Short answer. That means there exists an injection $\iota:C([0,1])\to L^\infty([0,1])$ and one can identify $C([0,1])$ with the image $\iota(C([0,1]))$, which is a subset of $L^\infty([0,1])$. Now a long one. We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially boun... (more) |
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| Edit | Post #290537 |
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| Edit | Post #290537 | Initial revision | — | about 1 year ago |
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What does it mean by saying that $C([0,1])$ is a subset of $L^\infty([0,1])$? >Question. What does it mean by saying that $C([0,1])$ is a subset of $L^\infty([0,1])$? Notes. This question might seem "obvious" to those experienced in analysis but can confuse beginners who interpret the statement "literally": an element in \( C([0,1]) \) is fundamentally different from an el... (more) |
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| Edit | Post #290527 |
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| Edit | Post #290528 |
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| Edit | Post #290527 |
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| Edit | Post #290528 |
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| Edit | Post #290528 | Initial revision | — | about 1 year ago |