Hot answers tagged

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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