# Tag Info

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Triangle inequality $$|\, ||f_n ||_2 -|| f ||_2\, |\leq ||f_n-f||_2$$

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since $\langle T(x+y),x+y\rangle =0$ that implies : \begin{eqnarray} \langle T(x+y),x+y\rangle &=&\langle Tx+Ty,x+y\rangle \\ &=&\langle Tx,x+y\rangle+\langle Ty,x+y\rangle\\ &=& \langle Tx,x\rangle+\langle Tx,y\rangle+\langle Ty,x\rangle+\langle Ty,y\rangle\\ \end{eqnarray} Then $$\langle T x,y\rangle +\langle Ty,x\rangle=0 \qquad ... 3 A counterexample for d=2: let \Omega be the disk \{x:\|x\|<\exp(-\exp(\pi))\}, and$$ f(x) = \sin \log \log \frac{1}{\|x\|} $$This function is in W_0^{1,2}(\Omega)\cap L^\infty(\Omega) (relevant calculations here) but has a discontinuity at 0, and moreover cannot be made continuous by redefining it on a set of measure zero. On the other ... 2 According to Daniel Fischer: The left hand side cries for an application of the Cauchy-Schwarz inequality. And according to siminore:$$ x=\sum_n x_n e_n = \sum_n \langle x,e_n \rangle e_n \quad ; \quad y=\sum_n y_n e_n = \sum_n \langle y,e_n \rangle e_n $$But we give it a twist:$$ x'=\sum_n |x_n| e_n = \sum_n \left|\langle x,e_n \rangle\right| e_n \quad ; ...

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Hah! This is actually a specific example of something in my research! (My work attacks a more general set of integral equations, in some sense.) Let's go for something nontrivial (unlike previous answers/comments). If you consider what I like to call a diagonal kernel, i.e. $g(x,t) = f(xt)$ for some $f$ and assume $g$ is real analytic, then this is very ...

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An easy solution can be obtained by making $g(x,t)$ degenerate: $$g(x,t)={1\over2}\exp(-bx^4)\exp(at^4-|t|)$$ which is susceptible to the generalization $$g(x,t)={1\over C}\exp(-bx^4)\exp(at^4)f(t)$$ where $f$ is integrable over $\mathbb R$ and $$\int_{-\infty}^{\infty}f(t)dt=C\neq0$$

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Well, $$||x||^{m^2}=(||x||^m)^m\le(K||x^m||)^m=K^m||x^m||^m\le K^mK||(x^m)^m||=K^{m+1}||x^{m^2}||.$$Now your argument for $n\le m$ also works for $n\le m^2$...

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Consider $u\in B(H_1\oplus H_2, H_1\oplus H_2)$ and since $u^*u$ is projection, then $u$ is partial isometry, which means that $uu^*u=u.$

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