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3

By definition $\|A\|=\sup_{\|x\|\leq 1} \|Ax\|$, and by Hahn-Banach theorem we can show $\forall y\in Y, \|y\|=\sup_{\|f\|\leq 1,f\in Y^*} \|f(y)\|$. Combining them, we have $\|A\|=\sup_{\|x\|\leq 1} \sup_{\|f\|\leq 1,f\in Y^*} \|f(Ax)\|$, which is what you want.

3

Suppose $T$ is a bounded operator on a Banach space $X$. $\lambda\in\rho(T)$ iff $T-\lambda I$ is a linear bijection. In that case, the inverse $(T-\lambda I)^{-1}$ is automatically continuous by the closed graph theorem. There are three basic things that can stand in the way of $T-\lambda I$ being invertible. $T-\lambda I$ is not injective. Equivalently, ...

2

If $g$ is an element of order $p$ (this always exists by Cauchy's Theorem), let $$Q=\frac1p\,\sum_{j=0}^{p-1} g^j.$$ Then $\text{tr}(Q)=1/p$ and $$Q^*=\frac1p\,\left(\sum_{j=0}^{p-1} g^j\right)^*=\frac1p\,\sum_{j=0}^{p-1} g^{-j} =\frac1p\,\sum_{j=0}^{p-1} g^{p-j}=\frac1p\,\sum_{k=1}^{p} g^k=\frac1p\,\sum_{k=0}^{p-1} g^k=Q.$$ Note that ...

2

As you have defined: $$p_nf = \sum_{k=1}^{n}\left[n\int_{\frac{k-1}{n}}^{\frac{k}{n}}f(y)dy\right]\chi_{[\frac{k-1}{n},\frac{k}{n}]}(x)$$ The linear operator $p_n$ is an orthogonal projection operator onto the linear span of the orthonormal set $\left\{\sqrt{n}\chi_{[\frac{k-1}{n},\frac{k}{n}]}\right\}_{k=1}^{n}$. Hence, $p_n^2=p_n$ and $\|p_nf\| \le ... 2 Actually convolution with$k$defines a bounded operator on$L^2(G)$if and only if$\hat k$is bounded. In this case the spectrum is exactly the essential range of$\hat k$. In fact if$m$is any bounded measurable function on$\hat G$then$m$defines a bounded operator on$L^2(G)$by $$\widehat{Tf}=m\hat f,$$whether or not there exists$k$with$m=\hat ...

2

Given such an operator $T$, note that for any polynomial $p$, you have $$T (p) = T (p \cdot 1) = p \cdot T (1)$$ Now use density of the polynomials to conclude $Tf = f \cdot T (1)$ for all $f \in C ([0,1])$, so that $T$ is a multiplication operator.

1

I am going to assume $\mathbf{A}$ is symmetric. Then it can be diagonalised by an orthogonal matrix $\mathbf{P}$ so that $\mathbf{A}=\mathbf{P^T B P}$, where $\mathbf{B}$ is diagonal. Substitute it into the integral and then make the substitution $\mathbf{u}=\mathbf{Pv}$. The Jacobian is $\mathbf{\det{P}}=1$. $$\int_{-\infty}^{\infty} \mathrm{d}v_1 ... 1 The operator -\Delta is densely defined on \mathcal{D}(-\Delta)=\mathcal{C}_{0}^{\infty}(0,\infty). If f,g \in \mathcal{D}(-\Delta), then the evaluation terms vanish when integrating by parts in the following:$$ (-\Delta f,g)=(f,-\Delta g) $$So -\Delta symmetric on its domain, which is enough to guarantee that -\Delta is ... 1 Indeed in 2-D, the only nontrivial nilpotent matrix is \left(\begin{array}{cc}0 & \epsilon\\ 0 & 0 \end{array}\right), \epsilon\neq0, (or it's transpose), and the equality can hold iff q=\frac{d_{ii}}{d_{jj}}. So this argument suggests it cannot be done. Next, why not suppose that N is upper triangular. Since N has all zeros on the ... 1 The answer to the question in the title is no. Let H=\ell^2(\mathbb N),$$ N=\{x\in H:\ \exists m:\ x(n)=0,\ \forall n\geq m\} $$and$$ M=\{\lambda z:\ \lambda\in\mathbb C\}, $$where$$ z=\left(1,\frac12,\frac13,\frac14,\ldots\right).  Then $N$ is dense, $M$ is closed, and $N\cap M=\{0\}$.

1

This is not an answer, just an elaboration of a comment. You have $(A-zI)^{-1} = (z({A \over z} -I))^{-1} = {1 \over z} ({A \over z} -I)^{-1} = -{1 \over z} (I-{A \over z} )^{-1}$. If $|z| > \|A\|$, we have $(A-zI)^{-1} = -{ 1\over z} (\sum_{k=0}^\infty ({A \over z})^k) = -{ 1\over z} (I+{A \over z}\sum_{k=0}^\infty ({A \over z})^k)$. Note that $\|{A ... 1 For$T$to have$\lambda$as an eigenvalue,$T-\lambda I$must be non-injective. For$\lambda$to be in the spectrum of$T$, it must only be non-invertible. These are equivalent when$T$is an operator on a finite-dimensional space, but not in general! For example, let$T$be the shift operator$(x_0,x_1,\dots) \mapsto (0,x_0,x_1,\dots)\$ on your favorite ...

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