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# Activity for Snoopy‭

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Edit Post #290571 Post edited:
8 days ago
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8 days ago
Edit Post #290642 Post edited:
3 months ago
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3 months ago
Edit Post #290642 Post edited:
3 months ago
Edit Post #290643 Post edited:
3 months ago
Edit Post #290643 Post edited:
3 months ago
Edit Post #290643 Post edited:
3 months ago
Edit Post #290643 Post edited:
3 months ago
Edit Post #290643 Post edited:
3 months ago
Edit Post #290642 Post edited:
3 months ago
Edit Post #290643 Initial revision 3 months ago
Answer 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...
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3 months ago
Edit Post #290642 Initial revision 3 months ago
Question 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...
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3 months ago
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3 months ago
Edit Post #290599 Post edited:
3 months ago
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Edit Post #290600 Initial revision 3 months ago
Answer 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) 3 months ago Edit Post #290599 Initial revision 3 months ago Question 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} $$Notes. This is an exercise in calculus. There are different ways to approach this problem. One may find the sum on the left explicitly with hyperbolic functions. One can also s... (more) 3 months ago Edit Post #290568 Post edited: 3 months ago Edit Post #290583 Post edited: 3 months ago Edit Post #290583 Initial revision 3 months ago Answer A: The derivatives of a function at a boundary point Your question is better phrased and answered in Whitney's paper: Analytic extensions of differentiable functions defined in closed sets. Here is an excerpt from the introduction section of the paper, which can be seen from the given AMS link: >Imagealttext (more) 3 months ago Comment Post #290576 https://en.wikipedia.org/wiki/Whitney_extension_theorem (more) 3 months ago Edit Post #290571 Post edited: 3 months ago Edit Post #290571 Post edited: 3 months ago Edit Post #290571 Initial revision 3 months ago Answer A: Using convexity in the proof of Hölder’s inequality The first question can be rephrased in a clear way as follows: > Let A and B be two positive real numbers. Show that the function F(t):=A^{p(1-t)}B^{qt} for t\in[0,1] is convex using the convexity of the exponential function. If one rewrites:$$ A^{p(1-t)}B^{qt} = e^{p(1-t)\log(A)+(q...
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3 months ago
Edit Post #290570 Initial revision 3 months ago
Question 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) 3 months ago Edit Post #290568 Post edited: 3 months ago Edit Post #290568 Post edited: 3 months ago Edit Post #290568 Post edited: 3 months ago Edit Post #290568 Initial revision 3 months ago Answer 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) 3 months ago Edit Post #290567 Post edited: 3 months ago Edit Post #290567 Initial revision 3 months ago Question 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) 3 months ago Edit Post #290538 Post edited: 3 months ago Edit Post #290538 Post edited: 3 months ago Edit Post #290538 Post edited: 3 months ago Edit Post #290538 Initial revision 3 months ago Answer A: What does it mean by saying that$C([0,1])$is a subset of$L^\infty([0,1])$? We say that a measurable function$f: [0,1] \to {\mathbf C}$is essentially bounded if there exists an M such that${|f(x)| \leq M}$for almost every$x\in[0,1]$, and define$\\\|f\\\|{L^\infty}$to be the least$M$that serves as such a bound. Let$\mathscr{L}^\infty([0,1])$denote the set of me... (more) 3 months ago Edit Post #290537 Post edited: 3 months ago Edit Post #290537 Initial revision 3 months ago Question 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 is an example of questions that are quite "obvious" to experienced readers in analysis but may be very confusing to beginners who take the statement "literally": an element in$C([0,1])\$ is a comp...
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3 months ago