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Activity for Snoopy‭

Type On... Excerpt Status Date
Edit Post #290571 Post edited:
8 days ago
Edit Post #290570 Post edited:
8 days ago
Edit Post #290642 Post edited:
3 months ago
Edit Post #290643 Post edited:
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
Edit Post #290600 Post edited:
3 months ago
Edit Post #290599 Post edited:
3 months ago
Edit Post #290600 Post edited:
3 months ago
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 $$ \begin{align} \sum{n=...
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3 months ago
Edit Post #290599 Initial revision 3 months ago
Question For any real number $m$, $ \left|\sum_{n=1}^{\infty}\frac{m}{n^2+m^2}\right|<\frac{\pi}{2} $
> Problem. Prove that for any real number $m$, $$ \left|\sum{n=1}^{\infty}\frac{m}{n^2+m^2}\right|<\frac{\pi}{2} $$ Notes. This is an exercise in calculus. There are different ways to approach this problem. One may find the sum on the left explicitly with hyperbolic functions. One can also s...
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3 months ago
Edit Post #290568 Post edited:
3 months ago
Edit Post #290583 Post edited:
3 months ago
Edit Post #290583 Initial revision 3 months ago
Answer A: The derivatives of a function at a boundary point
Your question is better phrased and answered in Whitney's paper: Analytic extensions of differentiable functions defined in closed sets. Here is an excerpt from the introduction section of the paper, which can be seen from the given AMS link: >Imagealttext
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3 months ago
Comment Post #290576 https://en.wikipedia.org/wiki/Whitney_extension_theorem
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3 months ago
Edit Post #290571 Post edited:
3 months ago
Edit Post #290571 Post edited:
3 months ago
Edit Post #290571 Initial revision 3 months ago
Answer A: Using convexity in the proof of Hölder’s inequality
The first question can be rephrased in a clear way as follows: > Let $A$ and $B$ be two positive real numbers. Show that the function $F(t):=A^{p(1-t)}B^{qt}$ for $t\in[0,1]$ is convex using the convexity of the exponential function. If one rewrites: $$ 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 ...
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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...
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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...
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3 months ago
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3 months ago
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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...
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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
Edit Post #290527 Post edited:
3 months ago