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 Apr 25 comment Let $f$ be analytic in $D(0,2)$. Assume that for all $n∈\mathbb{N}$ $\int_{ |z|=1} {f(z)\over(n+1)z−1}dz=0$. Prove that $f(z)=0$ for all $z∈D(0,2)$. @DerpMagoo You should get $f(\frac{1}{n+1})=0$ for all $n \in \mathbb{N}$. Also, note that $\frac{1}{n+1} \to 0 \in D(0,2)$ when $n \to \infty$ (so, you have sequence tending to element of $D(0,2)$ and on that sequence your function is zero, so...) Apr 25 comment Let $f$ be analytic in $D(0,2)$. Assume that for all $n∈\mathbb{N}$ $\int_{ |z|=1} {f(z)\over(n+1)z−1}dz=0$. Prove that $f(z)=0$ for all $z∈D(0,2)$. @DerpMagoo Yes. And if you take $a=\frac{1}{n+1}$ and look at your condition, you will get... Try it. Apr 25 comment Let $f$ be analytic in $D(0,2)$. Assume that for all $n∈\mathbb{N}$ $\int_{ |z|=1} {f(z)\over(n+1)z−1}dz=0$. Prove that $f(z)=0$ for all $z∈D(0,2)$. @DerpMagoo Do you know what is it Cauchy's integral formula? Apr 25 answered Let $f$ be analytic in $D(0,2)$. Assume that for all $n∈\mathbb{N}$ $\int_{ |z|=1} {f(z)\over(n+1)z−1}dz=0$. Prove that $f(z)=0$ for all $z∈D(0,2)$. Jan 9 awarded Yearling Nov 17 comment What does the theorem “If $\lim_{x\to a} f(x) = L$ and $f(x) = g(x)$ except at $a$ then $\lim_{x\to a} g(x) = L$” mean? It means that limit is not depending on value of function in that point where we are taking limit. That is, for $\lim_{x\to a} f(x)$ it is not important what it is $f(a)$ Oct 11 awarded Nice Answer Sep 5 comment Proving that an operator $T$ on a Hilbert space is compact @johny No. That is just arbitrary sequence of elements of our Hilbert space which is weakly convergent. I used that bounded operator is compact iff operator maps weakly convergent sequences into strongly convergent sequences. Sep 5 answered Proving that an operator $T$ on a Hilbert space is compact Sep 2 comment Why is $\overline{\operatorname{span}\{e_n\mid n\in\mathbb{Z}\}}=L^2(\mathbb{T})$? It's funny how you underline "Hint", but (at least on my monitor) it look like you overline "span". Jul 29 accepted Is it true $\lambda_i(f(A))=f(\lambda_i(A))$ for non-negative and non-decreasing function on $[0,\infty)$ Jul 28 comment Is it true $\lambda_i(f(A))=f(\lambda_i(A))$ for non-negative and non-decreasing function on $[0,\infty)$ First, thanks. I don't see immediately why $f(M_g) = M_{f \circ g}$? Also, I don't know this version of spectral theorem. Where can I find it? (I know variant: for self-adjoint $A$ there exist spectral measure $E$ such that $A=\int_{\sigma(A)} \lambda \, dE(\lambda)$) Jul 26 comment Is it true $\lambda_i(f(A))=f(\lambda_i(A))$ for non-negative and non-decreasing function on $[0,\infty)$ @PhoemueX Can you show me solution? Because I really don't know how to attack problem. Only I can see that from $\lambda_1(A) \geq \lambda_2(A) \geq \ldots \geq 0$ (if we arrange in that way) we get $f(\lambda_1(A)) \geq f(\lambda_2(A)) \geq \ldots \geq 0$ and I think that is very important. Jul 25 comment Evaluate the Integral: $\int(x^5+5^x)\ dx$ You can split in sum of two integrals. Jul 25 comment Is it true $\lambda_i(f(A))=f(\lambda_i(A))$ for non-negative and non-decreasing function on $[0,\infty)$ @PhoemueX I am mainly interested in eigenvalues (that is singular values (but that is the same here because $A$ is positive)). I would like (if that is possible) to see also why $\sigma(f(A)) \neq f(\sigma(A))$. Jul 25 revised Is it true $\lambda_i(f(A))=f(\lambda_i(A))$ for non-negative and non-decreasing function on $[0,\infty)$ edited title Jul 25 comment Is it true $\lambda_i(f(A))=f(\lambda_i(A))$ for non-negative and non-decreasing function on $[0,\infty)$ @DanielFischer Yes, you are right, I didn't thinking enough. And yes, we assume that $A$ is continuous linear operator. Jul 25 asked Is it true $\lambda_i(f(A))=f(\lambda_i(A))$ for non-negative and non-decreasing function on $[0,\infty)$ Jul 22 comment Integration of $\int \frac{(1 + x)\sin x}{(x^2 +2 x)\cos^2 x-(1 + x)\sin2x}dx$ Hm, just for note: derivative of $x^2+2x$ is $2(1+x)$ and derivative for $\cos^2 x$ is $-\sin 2x$. Jul 22 comment Serge Lang Never Explains Anything Round II @D_S I don't know this things. But if you changed $K'$ with $M$, then, do you need to change it everywhere? (you have in text $K'/k'$....)