# Tag Info

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 ...

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 ...

5

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

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

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 $k$.

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

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

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

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

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 $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$

3

If you have $$\alpha e^{x} + \beta e^{2x} + \gamma e^{3x} \equiv 0,$$ Then you can apply the derivative operator $D$ to obtain \begin{align} 0 & \equiv (D-2)(D-3)\{\alpha e^{x} + \beta e^{2x} + \gamma e^{3x}\} \\ & = (1-2)(1-3)\alpha e^{x}. \end{align} Therefore $\alpha=0$. Then you can apply $(D-1)(D-3)$ in order to ...

3

You can write the similarity as $NS=SM$. As $N$ and $M$ are normal, the Fuglede-Putnam theorem guarantees that $N^*S=SM^*$. Taking adjoints, $S^*N=MS^*$. Then $$S^*SM=S^*NS=MS^*S.$$ Using this identity repeteadly, $p (S^*S)M=Mp (S^*S )$ for all polynomials; taking limits, $f (S^*S)M=Mf (S^*S)$ for all continuous functions $f$. In particular, if ...

3

The function $x\mapsto d(p_0,x)$ is continuous on $X$, hence attains its minimum on $S$ since $S$ is compact.

3

You can construct the completion of a normed vector space $X$ by using a construction of the completion of a general metric space and then defining a normed vector space structure on that completion. But there's a better way, or at least a way that a lot of people might regard as simpler and more elegant: If $X$ is a normed vector space then the dual $X^*$ ...

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

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 ...

2

Consider the map $f$ defined by $x \mapsto \frac{Ax}{\sum_i (Ax)_i}$ defined on the (topological) disk $D$ that consists of vectors $x$ satifying $x_1 \ge 0, x_2 \ge 0, \ldots, x_n \ge 0, x_1 + x_2 + \ldots + x_n = 0$ (i.e., $D$ is the standard simplex in the positive octant). Then $$f : D \to D$$ is a continuous map of a closed disk to itself (this ...

2

I assume you want this $\forall K>0$, not all $t$. Since $f(t)^Tf(t)\geq0$, you can take $\beta=(\int_0^{\infty}f(t)^Tf(t)\,dt)^{1/2}.$

2

Consider the case $n=2$, and take $$T_k = \pmatrix{k^2 \sin(1/k) & k^2 \cos(1/k)\cr 0 & 1\cr}$$ Write $x \in \mathbb S^1$ as $\pmatrix{\cos(\theta)\cr \sin(\theta)}$, and note that $(T_k x)_1 = k^2 \sin(\theta + 1/k)$. If $\sin(\theta) = 0$ this is $\pm k + O(1)$, while if $\sin(\theta) \ne 0$ it is $k^2 \sin(\theta) + O(k)$. So $\|T_k x\| \to ... 2 The Spectral Theorem for$A$is given in terms of a Borel Spectral measure$E$$$Ax = \int_{-\infty}^{\infty}\lambda dE(\lambda)x,$$ and$x \in \mathcal{D}(A)$iff $$\int_{-\infty}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty.$$ The operator$e^{iA^2}is defined through the functional calculus as $$e^{iA^2}x = ... 2 If (X,d) denotes a metric space then we can construct a metric d' on X by stating:$$d'(x,y)=\min(d(x,y),1)This function can be shown to be a metric and induces the same topology as the original d. However every subset of X is bounded with respect to the constructed metric. 2 You want to show: \int_a^b \frac{1}{L} e^{-2 \pi inx/L}e^{2\pi imx/L}dx=0 \begin{align} <a_n|a_m> & = \int_a^b \frac{1}{L} e^{-2 \pi inx/L}e^{2\pi imx/L}dx \\ & = \frac{1}{L} \int_a^b e^{2 \pi i x(m-n)/L} dx\\ & = \frac{1}{L} \frac{L}{2 \pi i(m-n)} \left( e^{2 \pi i (m-n) b/L}-e^{2 \pi i(m-n)a/L}\right) \\ & = \frac{1}{2 \pi ... 2 Yes, it is correct. The operators T_N are of finite rank (hence compact) and converge to T. Thus, T is compact as well. 2 If X = \{p\} then X is connected. If X \neq \{p\} then X is not connected. This follows from the fact that in a metric space singletons are closed, together with the fact that in a connected space the only sets that are both open and closed are the empty set and the set itself. 2 A linear operator is continuous if and only if it is bounded. By one of the comments above, it is possible to show that the sequence a_k must be bounded (for otherwise, Tx for x\in\ell^1 would not itself be an element of \ell^1). Therefore, |a_k|\leq C for some C\geq 0. Next, recall that a linear operator T:\ell^1\rightarrow\ell^1 is ... 2 Defining \phi_\lambda(x)=\phi(\lambda x) for smooth \phi, the requirement is u(\phi_{\lambda})=\lambda^{-m-d}u(\phi)\quad\forall\phi\in C_0^{\infty}. $$If u happens to be a continuous function (and hence u(\phi)=\int u(x)\phi(x)), this is equivalent to what you wrote. 2 Let: B: C[a,b] \times C[a,b] \to \Bbb R be the (check that it is) bilinear form given by:$$B(x,y) = \int_a^b x(t)y(t) dt$$We have that:$$|B(x,y)| \le \int_a^b |x(t)||y(t)| dt \le (b-a)\|x\|_{\infty} \|y\|_{\infty} $$Hence, B is continuous and in particular \|B\| \le b-a. Now, f(x) = B(x,x) is the composition of two continuous functions, ... 2 If s \in S, then \mathcal{N}_s = \{ x \in H : \langle x,s \rangle = 0 \} is the null space of a continuous linear functional, which is the inverse image of \{0\} under this continuous linear functional. Hence \mathcal{N}_s is closed, as is the intersection$$ S^{\perp} = \bigcap_{s\in S}\mathcal{N}_s.$2 Take$l_\infty$and$S = \{e_k\}_{k \in \mathbb{N}}$. The$S$is closed, bounded, but not totally bounded since$\|e_i - e_j\| = 1$for all$i \neq j$(hence there can be no finite$\epsilon$-net for$\epsilon <1\$).

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