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"Nothing is more important than to see the sources of invention which, in my opinion, are more interesting than the inventions themselves."
--Gottfried von Leibniz (1646-1716)
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1h |
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Is this an integrable function? Yes, it is. See this post to see why $U(f)=0$. It is easy to show that $L(f)=0$. (You could also appeal to the theorem that a bounded function is Riemann integrable over $[a,b]$ if and only if its set of discontinuities in $[a,b]$ has measure zero.) |
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1h |
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Finding convergence of the next function: $f(x)=\sum_{n=1}^\infty\frac{(\ln n)^3}{n}x^n$ The Ratio Test, applied to $a_n={(\ln n)^3\over n}|x|^n$ returns $1\cdot |x|$. The series converges if this quantity is less than $1$ and diverges if this quantity is greater than $1$. |
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1h |
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Finding convergence of the next function: $f(x)=\sum_{n=1}^\infty\frac{(\ln n)^3}{n}x^n$ The Ratio Test, if that's a "trick". Also, check the points $x=1$ and $x=-1$ separately. |
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14h |
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construct a lebesgue integral function See this. |
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15h |
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Complex Analysis boundedness and limits What a strange problem ... |
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19h |
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If for every $x_n$ such that $x_n \rightarrow x$, there exists a $x_{n_k}$ such that $Tx_{n_k} = Tx$, is $T$ continuous? @ChristopherA.Wong Yes. But no subsequence of $(Tx_{n_k})$ could converge to $Tx$. |
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19h |
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If for every $x_n$ such that $x_n \rightarrow x$, there exists a $x_{n_k}$ such that $Tx_{n_k} = Tx$, is $T$ continuous? Yes. This follows from the fact that if $(x_n)$ converges to $x$ and $(Tx_n)$ does not converge to $Tx$, then there is an $\epsilon>0$ and a subsequence $(x_{n_k})$ of $(x_n)$ such that $\Vert Tx_{n_k}-Tx\Vert>\epsilon$ for all $k$. That this fact holds follows from the definition of convergence. |
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1d |
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Embedded Lp spaces To wit: for $\infty\ne q>p$, we have $\int |f|^p=\int_{[\,|f|\le 1\,]} |f|^p +\int_{[\,|f|> 1\,]} |f|^p\ \le\ \int_{[\,|f|\le 1\,]} 1 +\int_{[\,|f|> 1\,]} |f|^q$. So if the measure space is finite, then $f\in L_q$ implies $f\in L_p$. |
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1d |
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Embedded Lp spaces You don't need Holder to show merely set inclusion (Holder does, however, does give you the inequality between the norms). |
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1d |
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Embedded Lp spaces $P$ presumably is a probability measure. You could just use comparison tests. For the second inclusion, given $f\in L_2$, consider the sets where $|f|\le 1$ and where $|f|>1$. |
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1d |
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Find an interval of convergence and an explicit formula for $f(x)$ Consider the series $1+x^2+(x^2)^2+\cdots$ and $2x(1+x^2+(x^2)^2+\cdots)$. |
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1d |
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Is second derivative of a convex function convex? What would happen if you took two successive antiderivatives of a positive, non-convex function? |
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1d |
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Can't establish a lower bound on a supremum Oh, sorry, I didn't read carefully. What I mentioned in my first comment implies the norms are bounded below by $1$. What you have written above perplexes me as well ... |
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1d |
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Can't establish a lower bound on a supremum Just note: If $\alpha>k^{1/p}$, then $ \mu(\{x\in [0,1] : f_{k,j}(x) > \alpha\}) =0$. If $\alpha\le k^{1/p}$, then $ \mu(\{x\in [0,1] : f_{k,j}(x) > \alpha\}) =1/k$. |
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1d |
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$(x+2)\cos\frac1{x+2} - x\cos\frac1x > 2$ for $x\in[1,\infty)$ Indeed, $f''(x)=-(1/x^3)\cos(1/x)<0$ for $x>1$. So, the first derivative is decreasing on $[1,\infty)$. But $f'(x)=\cos(1/x)+(1/x)\sin(1/x)$ has limit $1$ at infinity, and $f'(1)>1$. So $f'(x)>1$ for $x\in[1,\infty)$. The Mean Value Theorem then gives the result. |
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1d |
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$(x+2)\cos\frac1{x+2} - x\cos\frac1x > 2$ for $x\in[1,\infty)$ My first inclination would be to find a bound for $f'$ and using the Mean Value Theorem. |
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1d |
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Improper Integral $\int_{1/e}^1 \frac{dx}{x\sqrt{\ln{(x)}}} $ The integral is not well-defined. $\ln(1/e)=-1$ ... |
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2d |
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About Uniform Convergence of $\sum_{n=1}^\infty\frac{\sin nx}{n}$ on $[0,2\pi]$ It's the Fourier series of the periodic extension of your function over $[0,2\pi]$. This is not continuous at even integer multiples of $\pi$. |
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2d |
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About Uniform Convergence of $\sum_{n=1}^\infty\frac{\sin nx}{n}$ on $[0,2\pi]$ No, see this and this. |
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2d |
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Improper Integral Question $ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}$ The integral is improper. If you just need to determine if the integral converges, you can use the comparison test ($0<{1\over 1+e^{2x}}<{1\over e^{2x}}$) to show it is convergent. |
