Tag Info

6

First note that the sequence is bounded: $$|\phi_n(f)|=n^{-1}\,\left|\sum_1^nf(j)\right|\leq\max\{|f(j)|:\ j=1,\ldots,n\}\leq\|f\|_\infty.$$ This shows that $\|\phi_n\|\leq1$ for all $n$, so the sequence $\{\phi_n\}$ lies in the unit ball of $(\ell_\infty)^*$. In the weak$^*$-topology, the unit ball of the dual is compact, and so every sequence within it ...

3

Because of the subspaces being selfadjoint, $B(H)V_1\subset V_2$ implies that $V_1B(H)\subset V_2$. If $V_1\ne0$, then $B(H)V_1B(H)\subset V_3$ contains all finite-rank operators, and thus $V_3$, being closed, contains the compact operators. If $V_3$ contains a non-compact operator, then the ideas in this answer show that $I\in V_7$ (I didn't count ...

2

Suppose you have a sequence which converges in $G$. The lower bound implies it converges in $F$ to the same limit. Suppose you have a sequence which ddoes not converge in $G$. The upper bound implies it does not converge in $F$. That's all you need, since metric spaces are sequential spaces.

2

There is nothing wrong about your argumentation. Here is a "real" counterexample: Set $$f(x) := 1_{[0,1/2]}(x) \qquad \text{and} \qquad g(x) := 1_{[1/2,1]}(x).$$ Then $$\|f\|_1 = \|g\|_1 = \frac{1}{2}$$ and therefore $$2 (\|f\|_1^2+ \|g\|_1^2) = 1.$$ On the other hand, it is not difficult to see that $$\|f+g\|_1 = \|f-g\|_1 = 1.$$ Hence, ...

1

The Fréchet derivative $\nabla f(x)$ of a function $f:X\to\mathbb{R}$ at $x$ is an element of $X^*$. And therefore, $\nabla f$ is a map from $X$ to $X^*$. The Lipschitz condition makes sense whenever there is a map between metric spaces; and both $X$ and $X^*$ have metrics induced by their norms. Explicitly, \| \nabla f(x) - \nabla f(y)\|_{X^*} ...

1

The symbols $\langle x^*,v_n\rangle$ just express $x^*(v_n)$, the functional $x^*$ evaluated at $v_n$. It is a common notation, inspired in the Hilbert space case, where the dual is the same original space.

1

Are you sure about your definition of $\mathcal H_0^{2,c}$? Shouldn't this be the class of $L^2$-bounded continuous martingales? This is a indeed a Hilbert space with the scalar product $\langle L,N\rangle =E[L_\infty N_\infty]$. This makes sense because $L$ and $N$ are uniformly integrable and therefore $L_\infty=\lim L_t$ exists by the martingale ...

Only top voted, non community-wiki answers of a minimum length are eligible