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5

Here's something to get you started. You know that $f(x)$ is a polynomial function, so it is also continuous. We also know that $$f(x)f(y)+2=f(x)+f(y)+f(xy)$$ if both $x$ and $y$ are zero. However, taking both $x$ and $y$ to approach zero, the equation is also true for either $x$ or $y$ or both being zero. Substitute $y=0$. Substituting and solving ...


5

Let our Banach space be $\ell^1$, which is the set of all $a \in \mathbb{R}^{\mathbb{N}}$ such that $\sum_n |a_n|<\infty$. The norm is the $L^1$ norm. For $i \in \mathbb{N}$, let $e_i \in \ell^1$ denote the $i^{th}$ standard basis vector, i.e, the sequence whose $i^{th}$ index is $1$, but all other indices are $0$. Consider the linear map $T: \ell^1 \to ...


4

Rather, each stage of the Cantor set's construction is comprised of a finite union of closed intervals, so is closed. The Cantor set, itself, is the intersection of these (countably-many) stages. As an intersection of closed sets, then, the Cantor set is closed.


3

It is well-known (and easy to see) that $c/Y$ is isomorphic to $c_0$. So we just need to renorm $c_0$ so that the natural map between them is an isometry. In fact, the norm $|||\cdot|||$ will do, where \begin{equation*}|||(x_n)|||=\frac{1}{2}\sup_{m,n}|x_n-x_m|.\end{equation*} Notice that for any $(x_n)\in c_0$ we have ...


3

I presume $\text{aff} A$ means the affine hull of $A$ (in some normed linear space), and $\text{ri} A$ is the relative interior of $A$ with respect to this affine hull. No, it's not true. In $\mathbb R^2$, let $A$ be a line segment, and let $B$ be the union of $A$ with some point not on $A$. Then $\text{ri}(A)$ is nonempty, but $\text{ri}(B)$ is empty.


3

What if you have eignnvectors of different eigenvalues? If $T(x) = \lambda x$ and $T(y) = \mu y$, then $T(x+y) = \lambda x + \mu y$, which is not a scalar multiple of $x + y$, unless $\lambda = \mu$.


3

The argument would work if $S_N$ was countably infinite, but the point is that this is the only way to get that $S_N$ is infinite. If we have assumed $\{\alpha : X^{\alpha} >0\}$ countable then each $S_n$ could easily be finite. But your issue is correct: $S_N$ needs only be infinite, nor necessarily uncountable.


3

To add the solution by @Rory Daulton: Set $y=0$ and you get $$f(x)f(0)+2=f(x)+2f(0)$$ With $f(0)=2$ you get $$f(x)=2f(0)-2$$ which contradicts $f^\prime(1)=3$. Thus $f(0)=1$. EDIT 1: Take $\partial_x$ from both sides in $$f(x)f(y)+2=f(x)+f(y)+f(xy)$$ which yields $$f^\prime(x)f(y)=f^\prime(x)+yf^\prime(xy)$$ Set $x=1$:$$3f(y)=3+yf^\prime(y)$$ Thus ...


3

Oh well, how silly of me not to have thought of the following argument sooner. In what follows, $ (e_{i})_{i \in I} $ is a self-adjoint approximate identity of $ A $ that is bounded in norm by $ 1 $. It is $ C^{*} $-folklore that such an approximate identity exists. Lemma. If $ (x_{i})_{i \in I} $ is a convergent net in $ A $ whose limit is $ x $, ...


3

Yes, it's complete. First, observe that your norm is comparable to the simpler norm $$ \|f\| = \sup_{n\in\mathbb{Z}} \int_n^{n+1}|f(y)|\,dy \tag{1} $$ Indeed, $\|f\|\le |f|\le 2\|f\|$ because every interval of length $1$ is contained in some interval of the form $[n,n+2]$. The space with $(1)$ is just the direct sum $\bigoplus_\infty X_n$ of Banach spaces ...


3

Hint Assume a point $(h,k)$ on the curve. Differentiate given equation to get $y^{'} = -\sqrt{\frac yx}$ Therefore equation of tangent using slope point form is $y-k = m(x-h)$ , where m is $y^{'}$ Put $x=0 , y=0$ for C and B , and then add the two


2

$$\langle w,w\rangle = 1$$ leads to $$\langle -u-v,-u-v\rangle =1.$$ Expand and simplify and you're done.


2

Using the spectral theorem: A selfadjoint operator $T \ne 0$ on a Hilbert space is compact iff $$ T = \lambda_1 E_{1} + \lambda_2 E_{2} + \cdots, $$ where $\{ E_{j} \}$ is a finite or countably infinite set of disjoint orthogonal projections onto finite-dimensional subspaces, and the sequence $\{ \lambda_{j} \}$, if infinite, converges to ...


2

You must show the set of Holder continuous functions is a countable union of nowhere dense sets in the metric space $C([0,1]).$ Here's a start: Show that $f(x)=x$ can be uniformly approximated by functions that are not Lipschitz. Hint: $\sqrt x$ is very close to $x$ on $[0,b]$ if $b$ is small enough. Added later: I'll write $C$ for $C([0,1])$ and $\|\,\|$ ...


2

The answer is "no". The unit sphere is norm closed in the unit ball under the (norm) subspace topology. But in an infinite dimensional normed space, the weak closure of the unit sphere is the unit ball. See this post for a proof of this. So, in the space $B(X, {\rm weak})$, the closure of the unit sphere is again all of $B_X$ (in general if $A$ is a ...


2

If $W(T)$ is real, $\langle Tx, x\rangle$ is real for all $x$. Thus $$\langle Tx, x\rangle = \overline{\langle Tx, x\rangle} = \langle x, Tx \rangle = \langle T^* x, x \rangle $$ so $\langle (T - T^*) x, x\rangle = 0$. Now use the polarization identity for the inner product $(x,y) \to \langle (T - T^*) x, y\rangle$.


2

Apart from trivial cases - all seminorms are identically $0$ - the two topologies don't coincide. A seminorm cannot distinguish $x$ and $-x$, so every neighbourhood of $x$ in $\mathcal{I}$ contains $-x$. For an $x$ and a seminorm $p$ with $p(x) \neq 0$, there is however a ball $\{ x : p(y-x) < \epsilon\}$ that doesn't contain $-x$. If you take the ...


2

Here is one particular example: If we consider Lebesgue measure, then we can consider $L^\infty([a,b],m)$, which has a norm given by $$\|\cdot\|_\infty \stackrel{\textrm{def}}{=} \operatorname{ess} \sup_{[a,b]} \cdot.$$ The essential supremum is defined as the smallest upper bound $a$ of $f$ such that $m\left(\{x : f(x) > a\}\right)) = 0$. This space is ...


2

Let's try a canonical test function, $u(x)=|x|^\alpha$ and see if some value of $\alpha$ works. Suppose $\Omega=B(0,1)$. Then $$ \sup_{x\in\Omega}|x|^\theta|u(x)|=1\tag{1} $$ as long as $\alpha+\theta\ge0$. $$ \int_\Omega|xu(x)|^{p^*}\,\mathrm{d}x =\omega_{N-1}\int_0^1r^{(1+\alpha)p^*}r^{N-1}\,\mathrm{d}r\tag{2} $$ which is finite when ...


2

Actually, from your relation, just apply "ri" to get $$\rm{ri\,conv\,rge}A=\rm{ri}(\rm{ri\,conv\,rge}A)\subset \rm{ri\,rge}A\subset\rm{ri\,conv\,rge}A,$$ from which $\rm{ri\,rge}A=\rm{ri\,conv\,rge}A$ is convex.


2

If you're going to apply multivariable calculus tools to the distance function, it's best to use the squared distance function: $$f(\omega)=\|x-\omega\|^2,\quad g(\omega)=\langle a,\omega\rangle$$ The minimum is attained in the same place, but this $f$ expands as inner product, allowing for simpler computations: $\nabla f(\omega) = 2(\omega-x)$. So, the ...


2

Here is one proof of $(\operatorname{aff} C - \operatorname{aff} C) \subset \operatorname{aff} (C - C)$. Note that $S$ is affine iff $S$ can be written as $\{x_0\}+L$ for some linear space $L$. Let $\operatorname{aff} C = \{x_0\} +L$. Then $\operatorname{aff} C - \operatorname{aff} C = \{x_0\} +L + \{-x_0\} +(-L) = L$, hence $\operatorname{aff} C - ...


2

Let $A= \{ (x,0) | x^2 \le 1\}$, $B=\{ (x,y) | x^2+y^2 \le 1 , y \ge 0\}$. Then $A \subset B$, but $\operatorname{ri} A = (-1,1) \times \{0\}$, $\operatorname{ri} B = \{ (x,y) | x^2+y^2 < 1 , y > 0\}$, hence $\operatorname{ri} A \cap \operatorname{ri} B = \emptyset$.


2

1) In that page the author is not claiming that $\eta_{\mathscr{I}}(A)$ exists (yet), but is rather discussing what properties it should have before constructing it. 2) Note that $\eta_{\mathscr{I}}(A)$ is an operator, not a function. The idea of functional calculus is that the map $f\longmapsto f(A)$ should be a $*$-homomorphism, i.e. it should preserve ...


2

What you said is correct. The following proof is essentially the same proof as you will give for your lemma HINT: For any $x\in X$ let $(x_n)$ be a sequence in $D$ such that $x_n\longrightarrow x$. Since $T$ is linear and bounded, we have \begin{equation} \|Tx_n-Tx_m\| = \|T(x_n-x_m)\|\leq\|T\|\|x_n-x_m\|. \end{equation} This shows that $(Tx_n)$ is a Cauchy ...


1

You only need to use that The sum and product of two such operators correspond to the sum and product of the corresponding functions. That said, for any operator $A$, the square of the operator $\eta_\mathscr I(A)$ equals to $\eta_\mathscr I^2(A)$ -- whatever it will mean --, but as a real (or complex) function, we have $\eta_\mathscr I^2=\eta_\mathscr ...


1

In your situation, you have $x_n \to x$. This can be proved by contradiction. Suppose $(x_n)$ does not converge to $x$. Then, there is $\varepsilon > 0$ and a subsequence $(x_{n_k})$ with $\|x_{n_k} - x\| \ge \varepsilon$ for all $k$. But this subsequence converges weakly to $x$ and, hence, is bounded. By compacity of $K$, a subsequence converges ...


1

The other implication is true as well, at least if you assume that $Y$ and $Z$ are closed in $X$. In this case, the projection maps are bounded (by the closed graph theorem) which is needed in the proof. Start with a bounded sequence $(x_n)=(y_n\oplus z_n)$ in $X=Y\oplus Z$. You are looking for a subsequence $(x_{n_k})$ such that the sequence ...


1

I think you are confused by Rudin's claim that the $f_k$ converge to a continuous function. What actually happens in Example 1.44 is that Rudin shows that the space $(C^0(\Omega),d)$ is complete, with $d$ the metric he defines in that example. The approach is the same as always: choose a Cauchy sequence, find some element in the space which deserves to be ...


1

Because $(\Delta f,f) \le -\lambda_1\|f\|^{2}$ for $f\in\mathcal{D}(-\Delta)$, then \begin{align} \frac{d}{dt}\|T(t)f\|^{2}& =(\Delta T(t)f,T(t)f)+(T(t)f,\Delta T(t)f) \\ & \le -2\lambda_1(T(t)f,T(t)f)= -2\lambda_1\|T(t)f\|^{2},\;\; f \in\mathcal{D}(-\Delta) \end{align} Hence, $$ \frac{d}{dt}\left( e^{2\lambda_1 ...



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