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13

You need to check if the functions are independent, as you said. A way to go about this, which that ties it in with things you likely know is to evaluate it at several points, as you did for $x=0$. You get one condition for $x=0$. You get another condition for $x=1$ and still another one for $x=2$. Each will allow more than one solution, but they'll ...


13

The answer to the question exactly as you asked it is yes; your space is isomorphic as a vector space, with no topology, to various Banach spaces. (See various comments for details.) Edit: The assertion that the answer is yes has met with vigorous disbelief. Also there's a technical point that I realized after some thought I simply didn't know how to do. ...


11

Write $$\alpha e^x + \beta e^{2x} + \gamma e^{3x} = 0$$ You can go ahead and cancel out a positive number like $e^x$ so: $$\alpha + \beta e^{x} + \gamma e^{2x} = 0$$ Suppose you have some solution for this with $\alpha$, $\beta$, $\gamma$ not all zero. Then, as you say $$ \alpha + \beta + \gamma = 0\qquad \qquad (1)$$ Because this must be true at $x = 0$ but ...


6

Hint: let $e^x=y$, $e^{2x}=y^2$, $e^{3x}=y^3$ you have: $\alpha y +\beta y^2+ \gamma y^3=0$ where the $0$ at RHS is the zero polynomial. Now: when a polynomial is the zero polynomial? In general: The $0$ at RHS is the neutral element for the sum of functions in the vector space, not simply the number $0$ and this means that it is the function ...


6

Ah, but how do you know $||f||=<f,f>$ defines a norm? That requires a proof too! Suppose $f\ne 0$. Then we can find $x \in [0,1]$ such that $f(x) \ne 0$. So $f(x)>0$ or $f(x)<0$. In either case, $f(x)^{2}>0$. But since $f$ is continuous at $x$, for any $\epsilon>0$ we can choose $\delta$ sufficiently small that if $|y-x|<\delta$, ...


6

Let $\{a_n\}$ be a sequence of positive real numbers that increases to $1$, with the property that the sequence of products $$ a_1,\ a_1a_2,\ a_1a_2a_3,\ a_1a_2a_3a_4,\ \ldots$$ converges to a positive value. It's not hard to write down a specific example. Let $\cal H = \ell^2(\mathbf R)$ and define $T : \cal H \to \cal H$ by $$T(x_1,x_2,x_3,\ldots) = (0, ...


6

There's a general framework that the Fourier transform fits into using Pontryagin duality and studying the characters of a locally compact abelian group, such as $\mathbb{R}$. The characters of $\mathbb{R}$ are exactly the maps $x \mapsto e^{itx}$, which is where the complex factor comes from. This has all sorts of wonderful consequences, like the fact that ...


5

Take $T_k x = k^2 x_1 -k x_2$. Then for any $x \neq 0$ we see that $|T_kx| \to \infty$. However, $T_k ({1 \over \sqrt{1 + k^2}}(1,k)) = 0$ for all $k$.


5

As a complement to the earlier (good) answer and comments: the space of all sequences (whether real or complex) arises in at least one fairly natural way, namely, as the continuous dual to the LF-space (strict inductive limit of Frechet spaces) $\mathbb R^\infty=\bigcup_n \mathbb R^n$, where $\mathbb R^n$ has its usual topology and is included in $\mathbb ...


5

Consider $f:\mathbb R\to\mathbb R $ defined by $f (x)=1$. The set $\{1\} $ is compact, but $f^{-1}(\{1\})=\mathbb R $ is not.


5

Let us assume that we are working with sequence spaces $\ell^p$ and $\ell^q$. Since the$$p=q=2$$case is taken care of already, assume$$p>2>q.$$Since $\ell^p\subset\ell^q$ for $q\le p$ as here, both sides of the desired inequality make sense. Start the indices of the sequences at zero, because we feel like it. So, an example to stretch things: ...


5

I'll assume $S$ not empty. Since the function $f\colon S\to\mathbb{R}$, $f(p)=d(p_0,p)$ is continuous, when $S$ is compact its image is compact, hence closed and bounded; therefore the image of $f$ contains its minimum. If $S$ is only assumed to be closed and bounded, but not compact, the statement is not generally true. Consider $X=\{0\}\cup (1,2]$, with ...


5

Hint: Use Wronskian and show that the Wronskian-Determinant does not vansish.


5

Just consider e.g. $$f(x) := \begin{cases} \frac{1}{\sqrt{x-5}}, & x \in (5,6), \\ 0, & \text{otherwise}. \end{cases}$$ Then $f \in L^1((1,\infty))$, but $f \notin L^2((1,\infty)$.


4

Since $K$ is compact, $0\in\sigma(K)$ (because $K$ is not invertible). Because $K$ is compact, any element of its spectrum has to be an eigenvalue. Now suppose that $\lambda\ne0$, and that $Ku=\lambda u$. This, with your concrete $k$, looks like $$ \lambda u(x)=\int_0^x y\,u(y)\,dy-x\,\int_1^x u(y)\,dy. $$ Since $u$ is integrable (otherwise $Ku$ makes no ...


4

This is not a complete answer. Rather, it is a suggestion on how to reduce the problem to a possibly simpler one. For each $c\in(-\pi/2,\pi/2)$, multiply each of your two identities by $e^{i c k - \epsilon k^2/2}$ with $\epsilon>0$, integrate in $k\in\mathbb R$, let $\epsilon\to0$, and then subtract one of the resulting identities from the other. ...


4

Notice that for each vector $x$, one has $$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$ Thus $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$


4

You have to prove $$ \forall x\in\mathbb{R}:\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\Leftrightarrow\alpha,\beta,\gamma=0, $$ but I think the quantifier applies only to the part on the left side of the $\Leftrightarrow$, like this: $$ \left(\forall x\in\mathbb{R}:\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\right) \Leftrightarrow\,\alpha,\beta,\gamma=0. $$ So ...


4

You need to show the three vectors are linearly independent. In this case I would use this trick; so that you don't need to worry about them being functions and the equality to hold for every value of $x$. If you consider $D: \mathcal{F} \rightarrow \mathcal{F}$, the derivative operator, is an endomorphism in $\mathcal F$ (i.e. a linear map from ...


4

We know that for $\|A\| < 1$, $(I-A)^{-1}$ is well-defined (prove this yourself if you have not done so yet) so we can talk about the inverse. Thus: $$ I = (I-A)(I-A)^{-1}.$$ Here is a hint: $$ 1 = \|I\| = \|(I-A)(I-A)^{-1}\|.$$ Try doing some basic norm manipulations to this. You need to increase the norm, not decrease it since you want a lower ...


4

A function is a machine, a machine that assigns to any value of some set $X$ a unique element $f(x)$ beloging to some set $Y$. So the proper notation would be $f:X\rightarrow Y:x\mapsto f(x)$. You read it as follows: "$f$ is a function from $X$ to $Y$ which assigns to any $x\in X$ a value $f(x)$ in $Y$". When you write $f$ you refer to the function, when you ...


4

I’d attack it much more directly. HINT: Suppose that $a=\langle a_n:n\in\Bbb Z^+\rangle\notin\ell_\infty$. Then $a$ has a subsequence $\langle a_{n_k}:k\in\Bbb Z^+\rangle$ such that $|a_{n_k}|\ge k$ for each $k\in\Bbb Z^+$. For each $k\in\Bbb Z^+$ let $$x_{n_k}=\frac1{ka_{n_k}}\;,$$ and let all the other terms of $x$ be $0$. Show that $x\in\ell_1$, ...


4

This holds for first countable or sequential spaces. It need not hold, even within functional analysis (the dual of $L^\infty([0,1])$ does not obey it, IIRC). For first countable spaces there is a simple theorem: $A \subseteq X$ is closed iff for all sequences $(x_n)$ that converge to $x$, if all $x_n$ are from $A$, then also $x \in A$. I.e. $A$ is closed ...


4

A dense set $D$ of $X$ is such that the closure of $D$ equals $X$. Or equivalently, every non-empty open set contains a pont of $D$. So the points of $D$ are in a sense "close" to all points of $X$, we can "approximate" points of $X$ by points in $D$. The name separable is somewhat unlucky (what can be separated, exactly?). It probably has an historic ...


4

You have \begin{align} a_1\sin(x+b_1)+a_2\sin(x+b_2)&=a_1\sin x\cos b_1+a_1\cos x\sin b_1+a_2\sin x\cos b_2+a_2\cos x\sin b_2\\ &=(a_1\cos b_1+a_2\cos b_2)\sin x+(a_1\sin b_1+a_2\sin b_2)\cos x. \end{align} So it is enough to show that $\alpha\sin x+\beta\cos x\in M_1$. For this, note that we can always find $t$ with $$ ...


3

An alternate approach is induction on $n=\dim(W)$. The base case $n=0$ is clear, so the hard part is the induction step. For this, it's enough to prove the following result: if $M$ is a closed subspace of $V$ and $x\in V,x\not\in M$, then $M+\mathbb{C}x$ is also a closed subspace. Indeed, by the Hahn-Banach theorem there is a continuous linear functional ...


3

Let $A \subseteq B(H)$ be any AW*-algebra that is not a von Neumann algebra. (Actually, we don't need a full AW*-algebra, see below.) Let $t \in A$ be any operator. By definition of AW*-algebra, every right-annihilator is generated by a projection. In particular, the right-annihilator of the singleton set $\{t\}$ is generated by a projection $q \in A$. ...


3

You can do it when the Banach space $E$ is separable (i.e. $\overline {\{x_n\}}=E$ for some sequence). Here a sketch of the way following the "Cours de Theorie de l'Approximation" of Professor P.J.Laurent (Grenoble University). Let $V$ a vectorial subspace of $E$ and $f\in V^*$ with $$||f||=\sup_{||x||\le 1;\space x\in V} |f(x)|$$ We have to show $\exists ...


3

This is studied in potential theory: the function $u$ is the Newtonian potential of $f$, $$u(x)=\int_{\mathbb{R}^n} K(x-y)f(y)\,dy$$ where $K(x)=c_n|x|^{2-n}$ for $n\ne 2$ and $K(x)=c_2\log|x|$ for $n=2$. In dimensions $n\ge 3$ the kernel $K$ decays at infinity, so $u(x)\to 0$ as $|x|\to\infty$ in this case, provided $f$ is reasonable (integrable and ...


3

The $\ell^p$ spaces are a special case of the $L^p$ spaces obtained by using the counting measure on the set of natural numbers. If you squint closely at the integral it looks like a sum or indeed as Forever Mozart points out: summation is just integration with the trivial measure on $\mathbb{N}$.



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