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3

The difference is due to the fact that Kreyszig consider the spectrum of unbounded operators. So for $\lambda$ in the continuous or residual spectrum, $R_\lambda$ exists (as an unbounded operator). The usual approach in functional analysis is for bounded operators, where you wouldn't allow for $R_\lambda$ to be unbounded, and so you would say it doesn't ...


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It's exactly as you say. If $g$ has two zeroes, then by Rolle there is a point where $g'=0$. This of course depends on $f$ being differentiable.


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Assume that $f(x)=x$ has two solutions $x\neq y$ so that $f(x) = x$ and $f(y) = y$. But then for some $\xi \in \left(x,\,y\right)$ the Mean Value Theorem gives $$f'(\xi)=\frac{f(x)-f(y)}{x-y}=\frac{x-y}{x-y}=1,$$ a contradiction. I'm assuming that you came across this question in Baby Rudin with the requirement that $f$ is differentiable. Both Rolle's and ...


3

You could try $\frac 2 \pi \arctan \lambda x$ for various $\lambda >0$. If you do not tell us what exactly you are looking for, we shall not be able to help.


2

Presumably you are asking about the absolute value of the diagonal terms, rather than the diagonal terms. In this case the answer is: Yes, since otherwise the diagonal elements would converge to zero. This would mean that $I$ is a norm limit of the finite-rank operators $$I_n = \sum_{k \le n} \lambda_k P_k,$$ where $\{P_k\}$ are the orthogonal projectors ...


1

Orthogonal: Let v be an inner product space and Let x,y€V. x is said to be **orthogonal** to y if <x,y>=0. Basis: A linearly independent subset S of a vector space V which spans the whole space V is called a Basis of the vector space.


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Orhtogonal: Let V be an inner product space and let x,y belongs to V. x is said to be Orthogonal to y if =0 Basis: let V be a vector space over a field. * A finite set of vectors v1,v2,.... ,vn in V is said to be Linearly Independent if α1v1+ α2v2+…..+ αnvn=0 => α1= α2= ……..= αn=0. * L(S)=V


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Usually the point spectrum consists only of the eigenvalues of $T$, that is, $$\sigma_p(T) = \{\lambda\in\mathbb{C}\ |\ T-\lambda I\text{ is not injective }\}.$$ Notice how this corresponds to the eigenvalues of $T$, since when $T-\lambda I$ is not injective, there exists some $x\neq 0$, such that $(T-\lambda I)x=0$, or $Tx=\lambda x$. My definition of the ...


1

If $|a_{i_i}| \leq 1$, then $$|b_i - a_{i_i}| = |(a_{i_i} + 1) - a_{i_i}| = |1| = 1.$$ If $|a_{i_i}| > 1$, then $$|b_i - a_{i_i}| = |0 - a_{i_i}| = |a_{i_i}| > 1.$$ Either way, we have $|b_i - a_{i_i}| \geq 1$.


1

Assume that for all $t$ we have $\sum _{r\geq 1} c_re_r(t)=\sum _{r\geq 1} c_r'e_r(t)$. Then we have $\langle \sum _{r\geq 1}c_re_r, e_i \rangle =c_i=\langle \sum _{r\geq 1}c_re_r, e_i \rangle =c_i'$ for all $i\geq 1$ (by linearity and orthogonality). This show uniqueness of coefficients.


1

If $\nabla \times \vec F=0$, then $\vec F=$ conservative if the domain is simply connected. The domain of the first example is not simply connected and thus if the curl of the vector is zero, one cannot conclude from that alone that the vector is conservative. The domains of the latter 2 examples are simply connected. Thus, if the curl is zero, then the ...



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