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For example, note that $f(2x) = f(x + x) = f(x) + f(x) = 2 f(x)$, and hence $f(x) = \frac{1}{2} f(2x)$, which means $f(\frac{1}{2} x) = \frac{1}{2} f(x)$. Similarly one can prove that, for every $q \in \mathbb{Q}$, we have $f(qx) = q f(x)$. Now suppose $\lambda \in \mathbb{R}$. Choose a sequence $q_n \in \mathbb{Q}$ such that $q_n \rightarrow \lambda$. This ...

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Since $$f(0)=f(0+0)=f(0)+f(0)=2f(0),$$ we have $f(0)=0$. It follows that $$f(-x)=f(-x)+f(x)-f(x)=f(-x+x)-f(x)=f(0)-f(x)=-f(x) \quad \forall x\in E.$$ Also for every $x\in E$ we have \begin{eqnarray} f(2x)&=&f(x)+f(x)=2f(x)\\ f(3x)&=&f(2x)+f(x)=3f(x)\\ &\vdots&\\ f(nx)&=&nx \quad \forall n\in \mathbb{N}. \end{eqnarray} ...

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Yes, the closure of a vector space is a vector space. Yes, $D(T)$ might not be a closed set. For example, polynomials form a non-closed subspace of $C[0,1]$ (the continuous functions on $[0,1]$).

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This is just a property of continuous functions. If $f(x)$ is continuous at $a$, and $\lim_{n\rightarrow\infty} a_n=a$, then $\lim_{n\rightarrow\infty} f(a_n)=f(\lim_{n\rightarrow\infty} a_n)=f(a)$. Be careful with your example as $\log(x)$ is not continuous at $x\leq 0$ so if $\lim_{n\rightarrow\infty} f(x)=0$ for some $x$, then the result doesn't ...

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We are talking about a flow field $v$ which does not change in time; but the field vectors change from point to point. Consider a fixed point $p$ within this flow field and a cube $C$ of side length $s\ll1$ centered at $p$. The net flux $\Phi$ of $v$ through the surface $\partial C$ of this cube represents the amount of fluid produced within $C$ per ...

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Imagine that the vector field in question describes the velocity of fluid at a given point in a giant tank of fluid. In this instance, a net positive divergence over a solid region means that there is fluid flowing out of that region or, equivalently, that fluid is being produced within the region, a 'source' if you like. A net negative divergence, on the ...

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First, suppose $T$ is bounded. Let $\{x_n\}$ be a sequence in $X$ converging to zero. Then, $\{\lVert x_n\rVert\}$ also converges to zero and hence is bounded (by some $M$). Thus, $\lVert Tx_n \rVert \leq \lVert T \rVert \lVert x_n \rVert \leq \lVert T \rVert M < \infty$. Conversely, suppose $T$ maps zero-convergent sequences to bounded sequences. If $T$ ...

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For your second question the above answer is good.... For the first one let $x,y \in \overline{D(T)}$, i.e., there exist sequences $(x_n),(y_n)$ in $D(T)$ such that $x_n \longrightarrow x$ and $y_n \longrightarrow y$, then $(x_n+y_n)$ is a sequence in $D(T)$ (as $D(T)$ is a vector space) such that $x_n+y_n \longrightarrow x+y$ and if $\alpha$ is any scalar ...

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If $1_A$ is the characteristic function of some set, then $$1_A \circ T=1_{T^{-1}(A)}$$ Therefore, if $f =\sum a_i 1_{A_i}$ you have $$f \circ T = \sum a_i ( 1_{A_i} \circ T)= \sum a_i 1_{T^{-1}(A_i)}$$ Thus $$\int\limits_X|f\circ T|^p\mathrm{d}m=\int\limits_X|\sum a_i 1_{T^{-1}(A_i)}|^p\mathrm{d}m=\int\limits_X|\sum a_i 1_{A_i}|^p\mathrm{d} T^* m$$

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It's not quite stated precisely. This should be more-or-less in Kato. The spectral projection for the isolated eigenvalue $\lambda$ is $$P = \dfrac{1}{2\pi i} \oint_\Gamma (z I-A)^{-1}\; dz$$ where $\Gamma$ is a small circle centred at $\lambda$. By assumption, this is a projection of finite rank. It is the limit (in operator norm) of the ...

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You can define $G$ as follows: if $x_n \in D(T)$ and $w \in W$ with $x_n \to w$, then $G(y) = \lim_{n \to \infty} T(x_n)$ (use boundedness of $T$ to show that this is well-defined). Then if $x_n \to w$ and $y_n \to v$, and $a,b$ are scalars, you want to show that $G(aw + bv) = a G(w) + b G(v)$. Well, what sequence (defined in terms of $x_n$, $y_n$, $a$, ...

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Suppose $$T(x_1)(t) = T(x_2)(t)$$for all $t \in[0,1]$. As you have already figured out that the range of the operator is the set of all continuously differentiable functions (this follows from Fundamental Theorem of Calculus). In light of your observation just differentiate the two sides of the equation to get $x_1(t) = x_2(t)$, ...

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