TITLE: $L_p$ norm of the Dirichlet Kernel
QUESTION [3 upvotes]: I am trying to show that $$ \left(\int_{-1/2}^{1/2} \left|\dfrac{\sin(\pi(2N+1)x)}{\sin(\pi x)}\right|^p dx \right)^{1/p}  \approx N^{\frac{p-1}{p}}$$ for all $1<p\leq \infty$ by using the fact that $|\sin(x)| \approx |x|$ if $x \in [-1/2, 1/2)$.
I have tried using the change $y=Nx$ and I end having the result $2N+1$. I do not know what I am doing wrong.
I would appreciate any help.

REPLY [0 votes]: Here is an alternative and direct proof:
Throughout this posting $1<p<\infty$ is fixed. Let $I=\big(-\tfrac12,\tfrac12\big]$ and $I_N=\big(-\tfrac{1}{(2N+1)},\tfrac{1}{2N+2}\big]$. The endpoints of $I_N$ are the two solutions to
$$D_N(x):=\frac{\sin((2N+1)\pi x)}{\sin\pi x}=0$$
closest to the origin. Recall that
$$\begin{matrix}
\frac{2}{\pi}\leq\frac{\sin t}{t}&\text{for all}&0\leq t\leq\frac{\pi}{2}\\
\Big|\frac{\sin t}{t}\Big|\leq1&\text{for all}& t\in\mathbb{R}
\end{matrix}$$
Over $I_N$, we have
$$\begin{align}
\int_{I_N}|D_N(x)|^p\,dx &=2\int^{\tfrac{1}{2N+1}}_0\Big|\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}\Big|^p\,dx\\
&\leq 2(\pi(2N+1)/2)^p\int^{\tfrac{1}{2N+1}}_0\Big|\frac{\sin((2N+1)\pi x)}{(2N+1)\pi x}\Big|^p\,dx\\
&\leq 2\Big(\frac{\pi}{2}\Big)^p(2N+1)^{p-1}
\end{align}$$
and
$$\begin{align}
\int_{I_N}|D_N(x)|^p\,dx &\geq 2(2N+1)^p\int^{\tfrac{1}{2N+1}}_0\Big|\frac{\sin((2N+1)\pi x)}{(2N+1)\pi x}\Big|^p\,dx\\
&\geq2(2N+1)^p\int^{\tfrac{1}{2(2N+1)}}_0\Big|\frac{\sin((2N+1)\pi x)}{(2N+1)\pi x}\Big|^p\,dx\\
&\geq\Big(\frac{2}{\pi}\Big)^p(2N+1)^{p-1}
\end{align}$$
Over $I\setminus I_N$ we have
$$\begin{align}
\int_{I\setminus I_N}|D_N(x)|^p\,dx&=2\int^{\tfrac12}_{\tfrac{1}{2N+1}}\Big|\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}\Big|^p\,dx\\
&\leq2\big(\frac{\pi}{2}\big)^p\int^{\tfrac{1}{2}}_{\tfrac{1}{2N+1}}\Big|\frac{\sin((2N+1)\pi x)}{\pi x}\Big|^p\,dx\\
&=2\big(\frac{\pi}{2}\big)^p(2N+1)^{p-1} \int^{\tfrac{(2N+1)\pi}{2}}_{\pi}\Big|\frac{\sin y}{y}\Big|^p\,dy\\
&\leq 2\big(\frac{\pi}{2}\big)^p(2N+1)^{p-1}\int^\infty_{\pi}\Big|\frac{\sin y}{y}\Big|^p\,dy\\
&\leq \frac{\pi(2N+1)^{p-1}}{2^{p-1}(p-1)}
\end{align}
$$
Putting things together, we obtain
$$
\frac{2}{\pi}(2N+1)^{1-\tfrac1p}\leq\|D_N\|_p\leq2^{1/p}\Big(\frac{\pi}{2}+\Big(\frac{\pi}{2(p-1)}\Big)^{1/p} \Big)(2N+1)^{1-\tfrac1p}
$$
From this, we obtain constants $C_p,c_p>0$ such that
$$c_pN^{1-\tfrac1p}\leq \|D_N\|_p\leq C_p N^{1-\tfrac1p}$$
for all $N$. This may not be the best constants however.


TITLE: Defining a line
QUESTION [2 upvotes]: I've read this:
To describe a line, we needed a point $b$ and a vector $v$ along the line. We could also start with two points $b$ and $a$ and take $v=a−b$.
In the first sentence, why do we need a point $b$ and a vector $v$ along the line in order to define it? Isn't a vector $v$ along it just enough? I see we don't need point $b$ with it . Am I right?

REPLY [0 votes]: A regular old vector simply just says "go right by x, up by y, and forwards by z". It's kind of like the gradient in y = mx + c, and is not to be confused with a position vector, which says "go from the origin to this point in a straight line".
We often combine a vector with a variable coefficient, such as $ t(1, 3, 5) $ in order to make it extend infinitely in both directions. However, it still has no set point. It's like a ruler , with a physically fixed angle of course, that you can place anywhere on a piece of paper to draw a line.
However, it has no definite equation until we introduce a point to create a vector equation of a line, where we say "it goes through this point, with this gradient". The same goes for y = mx + c when you have both a co-ordinate and the gradient.
One thing to note is that many fixed positions will create the same line, since the line extends infinitely, including negatively. If our equation is: $ r $1$ = (1, 2, 3) + t(2, 5, 10)$, this is equal to $ r $1$ = (3, 7, 13) + t(2, 5, 10)$. If our variable coefficient was to have a value, the specific fixed position would matter.


TITLE: Proving an operator is hermitian
QUESTION [3 upvotes]: I am supposed to prove an operator G is hermitian but I seem to be proving it isn't. G is defined as
$$G = -i(O-O^t)$$
where $O$ is some linear operator. If I take the transpose of the whole thing I get
$$G^t = -i(O^t - O),$$
which is $-G$. Am I missing something?

REPLY [3 votes]: The hermitian adjoint is not merely the transpose of an operator; it is the complex conjugate of the transpose; that is, for complex matrices $A$,
$A^\dagger = \overline{A^T} = (\bar A)^T; \tag 1$
so assuming $O$ is a real matrix, and
$G = -i(O - O^T), \tag 2$
we have
$G^\dagger = \overline{(-i(O - O^T))^T} = \overline{-i(O^T - O)}$
$= i(O^T - O) = -i(O - O^T) = G; \tag 3$
$G$ is indeed self-adjoint.


text:
/-
  Presheaf of rings.

  https://stacks.math.columbia.edu/tag/006N

  Author: Ramon Fernandez Mir
-/

import sheaves.presheaf

universes u v

-- Definition of a presheaf of rings.

structure presheaf_of_rings (α : Type u) [topological_space α]
extends presheaf α :=
(Fring           : ∀ (U), comm_ring (F U))
(res_is_ring_hom : ∀ (U V) (HVU : V ⊆ U), is_ring_hom (res U V HVU))

instance {α : Type u} [topological_space α]
: has_coe (presheaf_of_rings α) (presheaf α)
:= ⟨λ F, F.to_presheaf⟩

attribute [instance] presheaf_of_rings.Fring
attribute [instance] presheaf_of_rings.res_is_ring_hom

instance presheaf_of_rings.comm_ring {α : Type u} [topological_space α]
  (F : presheaf_of_rings α) (U : topological_space.opens α) : comm_ring (F U) :=
F.Fring U

namespace presheaf_of_rings

variables {α : Type u} {β : Type v} [topological_space α] [topological_space β]

-- Morphism of presheaf of rings.

structure morphism (F G : presheaf_of_rings α)
extends presheaf.morphism F.to_presheaf G.to_presheaf :=
(ring_homs : ∀ (U), is_ring_hom (map U))

local infix `⟶`:80 := morphism

def identity (F : presheaf_of_rings α) : F ⟶ F :=
{ ring_homs := λ U, is_ring_hom.id,
  ..presheaf.id F.to_presheaf }

-- Isomorphic presheaves of rings.

local infix `⊚`:80 := presheaf.comp

structure iso (F G : presheaf_of_rings α) :=
(mor : F ⟶ G)
(inv : G ⟶ F)
(mor_inv_id : mor.to_morphism ⊚ inv.to_morphism = presheaf.id F.to_presheaf)
(inv_mor_id : inv.to_morphism ⊚ mor.to_morphism = presheaf.id G.to_presheaf)

end presheaf_of_rings

subset_name: curated


TITLE: $\mathcal{E}(-,E)○A$ has a left adjoint
QUESTION [2 upvotes]: Theorem: If $A:\mathbf{C} \rightarrow \mathcal{E}$ is a functor from a small category $\mathbf{C}$ to a cocomplete category
$\mathcal{E}$, the functor $R$ from $\mathcal{E}$ to presheaves given
by, $$R(E):C \rightarrow Hom_{\mathcal{E}}(A(C),E)$$ has a left
adjoint $L:\hat{\mathbf{C}} \rightarrow \mathcal{E}$ defined for each
presheaf $P$ in $\hat{\mathbf{C}}$ as the colimit $$L(P)=colim(\int P \xrightarrow{\pi_{p}} \hat{\mathbf{C}} \xrightarrow{A} \mathcal{E}).$$

I'm trying to understand this theorem. What is the intuitive idea behind it?

REPLY [1 votes]: A few things: this left adjoint is, like any left adjoint, cocontinuous. Less obviously, $L$ restricts to $A$ itself, upon composing with the Yoneda embedding. This requires showing that any object is the colimit of its own representable functor's category of elements, a good thing to prove for yourself if you don't already know it. Finally, $L$ is the only possible cocontinuous functor extending $A$ in this way, which follows from the fact that every presheaf is the colimit of representables indexed by its category of elements. This is a harder but extremely important result (generalized by the theorem you've stated) sometimes called the co-Yoneda lemma.
In short, this result extends to show that cocontinuous functors out of $\widehat{\mathbf C}$ are "the same thing" as functors out of $\mathbf C$, precisely, the categories thereof are equivalent. This means the presheaf category is the free cocompletion of the category, in a way closely analogous to any free construction in classical abstract algebra--except that the presheaf category is large in size compared to the original category.


TITLE: simple random walk question
QUESTION [0 upvotes]: A simple random walk given as $S(n) = 0,1, 2, 3, ..., n$ representing the sum of i.i.d random variables, with the following:
$$P(Y_i = -1)=p, P(Y_i = 0)=1-p-q, P(Y_i=1)=q$$
Find $P[S(3)=0]$ when $p=0.15, q = 0.45$.
Given that we have a sum of i.i.d random variables, then $P(Y_3 = 0) = (1-p-q)^3=(1-0.15-0.45)^3=0.40^3$
First time working with random walks, I'm just basing this based on the i.i.d random variables. Have I correctly approached this?
We find that there are 9 combinations
$$\begin{align}(0,0,0) &= 0.40^3 \\ (0,1,0) &= 0.40^2\cdot0.15 \\(1,0,0) &= 0.40^2\cdot0.15 \\(1,1,0) &= 0.40\cdot0.15^2 \\(0,-1,0) &= 0.40^2\cdot0.45 \\(-1,0,0) &= 0.40^2\cdot0.45 \\(-1,-1,0) &= 0.40\cdot0.45^2 \\(1,-1,0) &= 0.40\cdot0.15\cdot0.45 \\(-1,1,0) &= 0.40\cdot0.15\cdot0.45 \\\end{align}$$
$$P[S(3)=0]=0.4^3+0.4^2\cdot \:0.15+0.4^2\cdot \:0.15+0.4\cdot \:0.15^2+0.4^2\cdot \:0.45+0.4^2\cdot \:0.45+0.4\cdot \:0.45^2+0.4\cdot \:0.15\cdot \:0.45+0.4\cdot \:0.15\cdot \:0.45 = 0.40$$

REPLY [3 votes]: Expanding my comments into a partial answer:
First, it looks like you're making an important mistake: I think you're conflating $S(3)$, which is the position of the random walk after 3 steps, and $Y_3$, which is the third step itself.
In your amended question, you worked hard to compute $\mathbb P(Y_3 = 0)$. I can confirm that you have correctly computed the probability that the third step is $0$, because you have exhausted all possibilities for how that might occur. Note that $\mathbb P(Y_3 = 0)$ is given to be $1 - p - q = 1 - 0.15 - 0.45 = 0.40$. So, the good news is that your calculation is correct -- but the bad news is that it's not useful, because it's not what you were asked to find (and it's a consequence of things you already knew without the laborious computation).
Instead, you were asked to find the probability that the walk was at position $0$ after three steps. This happens in three of the paths that you evaluated above -- specifically, $(1, -1, 0)$, $(-1, 1, 0)$, and $(0, 0, 0)$. However, there are others that you didn't consider that are relevant, such as $(1, 0, -1)$. Your task is to find all the triples that sum to $0$.
This can be daunting at first, but it might be easier if you group similar cases together. For instance, you already found the probabilities of $(1, -1, 0)$ and $(-1, 1, 0)$ to be the same; note that this is also the same as the probability of $(1, 0, -1)$, as well as several other paths that are unstated so far. So, you might consider grouping them by type and then counting how many of them there are.
Regarding your latest question in the comments: yes, directly counting the cases (as you've started to do) becomes considerably more complicated when you proceed to 4 steps. However, if you consider the grouping approach I outlined, the types don't get too much worse than they already are.


TITLE: Extracting motion data from a list of coordinates
QUESTION [0 upvotes]: Cross posted from stackoverflow at the suggestion of a commenter.
I have a series of CSV files of timestamped coordinates (X, Y, and Z in mm). What would be the simplest way to extract motion data from them?
Measurables
The information I'd like to extract includes the following:

Number of direction changes
Initial acceleration of the first and last movements
Bearing (angle) of these movements
Average speed whilst non-stationary

Ideally, I'd eventually like to be able to categorise patterns of motion, so bonus points for anyone who can suggest a way of doing this. It strikes me that one way I could do this would be to generate pictures/videos of the motion from the coordinates and ask humans to categorise them - suggestions as to how I'd do this are very welcome.
Noise
A complication is the fact that the readings are polluted with noise. In order to overcome this, each recording is prefaced with at least 20 seconds of stillness which can serve as a sort of "noise profile". I'm not sure how to implement this though.
Specifics
If it helps, the motion being recorded is that of a persons hand during a simple grabbing task. The data is generated using a magnetic motion tracker attached to the wrist. Also, I'm using C#, but I guess the maths is language agnostic.

Magnetic tracker spec: http://www.ascension-tech.com/realtime/RTminiBIRD500_800.php
Sample data file: http://tdwright.co.uk/sample.csv

Bounty
There is a bounty open on the original question, so feel free to post your answers there as well as here.

REPLY [0 votes]: If I were modeling this, I would treat the initial point in the file as the displacement relative to the some fixed point and each subsequent point as a displacement from the previous point. In this way, you effectively end up with a sequence of vectors that can be reasoned over using standard methods from vector analysis. That is, the points in the file $\{p_0, p_1, \dots p_n\}$ yield the sequence of vectors $\{v_0 = p_1, v_1 = p_1 - p_0, \dots v_n = p_n - p_{n-1} \}$. Using these vectors you can then find the angle between them, determine the average speed, etc, according to elementary formula. As for dealing with noise in your data, I can't comment much on that other than to note that it might be helpful to introduce a notion of tolerance similar to that which is introduced to determine whether two floating point numbers are "equal". If the magntitude of displacement between point $p_i$ and $p_j$ is less than some tolerance value $\epsilon$ and/or change in direction also falls within a certain level of tolerance then perhaps it can be ignored. There are of course much more sophisticated statistical techniques that can be employed that I'm not really qualified to explain.


TITLE: A step in verifying a stopping time.
QUESTION [0 upvotes]: Suppose $X$ is a cadlag process adapted to $\{\mathcal F_t\}$ and $H$ is a closed set.Verify $\sigma_H\triangleq\inf\{t\ge0:X_t(\omega)\in H\}$ is a stopping time .
The first step is:
$$\{\sigma_H\le t\}=\{X_0\in H\}\cup\{X_s\in H\text{ or }X_{s-}\in H \text{ for some }s\in\color{red}{(}0,t]\}$$
I wonder if I can omit the "$\text{ or }X_{s-}\in H$"?i.e.
$$\{\sigma_H\le t\}=\{X_0\in H\}\cup\{X_s\in H\text{ for some }s\in\color{red}{(}0,t]\}$$

REPLY [0 votes]: The equality
$$\{\sigma_H \leq t\} = \{X_0 \in H\} \cup \{X_s \in H \, \text{or} \, X_{s-} \in H \, \text{for some} \, s \in (0,t]\} \tag{1}$$
does not hold. Consider the following counter example:
Consider the (deterministic) cádlág process $$X_t := \begin{cases} t & t<1 \\ 0 & t \geq 1. \end{cases}$$ and set $H:=\{1\}$. Then it follows easily from the definition of $\sigma_H$ that $\sigma_H = \infty$ and therefore
$$\{\sigma_H \leq t\} = \emptyset$$
for all $t \geq 0$. On the other hand, the right-hand side of $(1)$ does not equal $\emptyset$ for $t \geq 1$; this follows from the fact that $X_{1-} = 1 \in H$.


TITLE: MathSciNet review alert?
QUESTION [26 upvotes]: Other than reading MathSciNet regularly, is there a way to get notified when one of my papers gets a review on MathSciNet?

REPLY [18 votes]: Though only semi-related to the question (couldn't submit it as a comment), I may take the liberty to add that for zbMATH subscribers there is an Atom Feed providing such regular updates for, e.g., subjects or authors. In your case, it would be sufficient to check your author ID at your profile https://zbmath.org/authors/?q=P%C3%A1lv%C3%B6lgyi+D%C3%B6m%C3%B6t%C3%B6r (i.e., Author ID palvolgyi.domotor) - and to use it to subscribe for https://zbmath.org/atom/ai/palvolgyi.domotor
with you favourite news reader.
Olaf Teschke, Managing Editor, zbMATH


TITLE: How to handle sums in Tait's reducibility proof of strong normalisation?
QUESTION [7 upvotes]: I've been reading Girard et al's 'Proofs and Types', which in Chapter 6 presents a proof of strong normalisation for the simply typed lambda calculus with products and base types. The proof is based on Tait's method of defining a set of 'reducible terms for type $T$' by induction on the type $T$. For example a term $t$ is reducible for type $A\to B$ if for all reducible $u$ for type $A$, $t\,u$ is reducible for type $B$.
However I don't see how to extend this proof to (binary) sums, which are covered in a later chapter, because the elimination for sums involves a 'parasitic' type which prevents one from defining reducible terms of type $A+B$ via the elimination by induction on $A+B$.
How (if at all) does this proof style extend to sums? I can't find a good reference that presents this.

REPLY [5 votes]: Let me add to the above (perfectly correct) answers that there is a more general perspective, which works for all positive connectives, of which sums are a special case.
The fundamental insight is that reducibility actually arises from a duality between terms and contexts, which is itself parametric in a set of terms closed under certain operations (such as $\beta$-expansion, but I think something trickier must be used for strong normalization), called a pole.  Given a pole $P$, one defines
$$\mathsf C\perp t\quad\text{just if}\quad\mathsf C[t]\in P$$
where $\mathsf C$ is a context (a term with a hole denoted by $\bullet$) and $t$ a term. The idea is that $P$ is a set of terms which "behave well" (e.g. they have a normal form), so $\mathsf C\perp t$ means that the context $\mathsf C$ and the term $t$ "interact well" with respect to the property expressed by $P$.
Given a set $S$ of contexts, one may define a set of terms
$$S^\bot:=\{t\mathrm{|}\forall\mathsf C\in S,\,\mathsf C\perp t\}.$$
A similar definition induces a set of contexts from a set of terms $T$, which we still denote by $T^\bot$.
We may then define the interpretation of a type $A$, which is a set of contexts denoted by $|A|$, by induction on $A$:
$$
\begin{array}{rcl}
|X| &:=& \{\bullet\} \\
|A\to B| &:=& \{\mathsf C[\bullet\,u]\mathrel{|} u\in|A|^\bot,\ \mathsf C\in|B|\} \\
|A\times B| &:=& \{\mathsf C[\pi_1\bullet]\mathrel{|}\mathsf C\in|A|\}\cup\{\mathsf C[\pi_2\bullet]\mathrel{|}\mathsf C\in|B|\}\\
|A+B| &:=& \{\mathsf C\mathrel{|}\mathsf C[\mathsf{inl}\,\bullet]\in |A|^{\bot\bot}\text{ and }\mathsf C[\mathsf{inr}\,\bullet]\in|B|^{\bot\bot}\}
\end{array}
$$
Finally, we let
$$\|A\|:=|A|^\bot,$$
which is a set of terms. This is the "reducibility predicate" on $A$.
One then goes on to prove the so-called adequacy theorem: if
$$x_1:C_1,\ldots,x_n:C_n\vdash t:A$$
is derivable, then for all $u_1,\ldots,u_n$, $u_i\in\|C_i\|$ implies $t[u_1/x_1,\ldots,u_n/x_n]\in\|A\|$. This is proved by induction on the last rule of the derivation, using the properties of the pole (closure under $\beta$-expansion or something like that).
Now, if things are setup correctly, one also has $x\in\|A\|$ as well as $\|A\|\subseteq P$ for every variable $x$ and every type $A$, which implies that every term belongs to the pole by adequacy, i.e., every term "behaves well".
I am sure that this works for weak normalization, i.e., one may take $P$ to be the set of weakly normalizing terms (which is closed under $\beta$-expansion) and immediately obtain weak normalization of the simply-typed $\lambda$-calculus with sums. For strong normalization, some kind of hack is needed, because $\beta$-expanding a strongly normalizable term does not necessarily yield a strongly normalizable term, but I'm sure that a suitable definition will make things work.
This "reducibility by duality" may be found already in Girard's Linear logic paper (1987), although I'm not sure he is the one who introduced it. It has been used by many authors since then and of course it easily extends to higher order quantification. In particular, the above presentation is adapted from Krivine, who uses it as the basis of his classical realizablity theory (see his many papers on the subject, all available on his web page).


TITLE: How to prove this set equation?
QUESTION [0 upvotes]: I started as
$$\bar{X_1}\cup(\bar{X_2}\setminus(X_3\cap\bar{X_1})=\bar{X_1}\cup(\bar{X_2}\setminus X_3)\cup (\bar{X_2}\setminus \bar{X_1})=\bar{X_1}\cup(\bar{X_2}\setminus X_3)\cup \bar{X_2}=\bar{X_1}\cup \bar{X_2}$$
To simplify the left side, but I do not know how to proceed further with the proof itself :(

REPLY [0 votes]: We can proceed with $\to$ first and then $←$.
For $→$ : assume that the equality holds and assume for contradiction that $X_1 ∩ X_2 ∩ X_3 \ne \emptyset$.
This means that, for some $z : z ∈ X_1,X_2,X_3$, and this in turn implies that:

$z ∈ X_3$ and $z ∉ \bar{X_1} ∪ \bar{X_2}$.

But thus : $z ∈ X_3 \cup \bar{X_1} \cup \bar{X_2}$.
In conclusion, we have that, for some $z : z ∉ \bar{X_1} ∪ \bar{X_2}$ and $z ∈ X_3 \cup \bar{X_1} \cup \bar{X_2}$, contradicting the fact that the two are equals.

Similarly for $←$ : assume that $X_1 ∩ X_2 ∩ X_3 = \emptyset$ and assume for contradiction that $\bar{X_1} ∪ \bar{X_2} \ne X_3 \cup \bar{X_1} \cup \bar{X_2}$.
This means that there is a $z$ such that $z ∈ X_3 \cup \bar{X_1} \cup \bar{X_2}$ and $z ∉ \bar{X_1} ∪ \bar{X_2}$.
But $z ∉ \bar{X_1} ∪ \bar{X_2}$ means that $z \in X_1$ and $z \in X_2$.
And $z ∉ \bar{X_1} ∪ \bar{X_2}$, with the fact that $z ∈ X_3 \cup \bar{X_1} \cup \bar{X_2}$, means that $z \in X_3$.
Thus :

$z \in X_1,X_2,X_3$,

contradicting the fact that $X_1 ∩ X_2 ∩ X_3 = \emptyset$.


TITLE: Basis of solution sets
QUESTION [1 upvotes]: I know that the collection of all solutions to $\sum_{i=0}^nA_iy^{(i)}(t)=0$ form a vector space. But in which way can one find out its basis? Of course I already learned what the basis is. But the key question is, how to prove that it is?

REPLY [1 votes]: Let's look at an example, $$y''+y'-2y=0$$ It seems that you know that $y_1(t)=e^t$ and $y_2(t)=e^{-2t}$ form a basis for the vector space of all solutions, but you want to know why.
First, let's prove that these two solutions are linearly independent. Suppose $$Ae^t+Be^{-2t}=0$$ identically (that is, for all $t$). Let $t=0$; you get $A+B=0$, so $B=-A$, so $$Ae^t-Ae^{-2t}=0$$ so $A=0$ or else $e^t-e^{-2t}$ is identically zero. The latter is nonsense ($e^t>1>e^{-2t}$ for $t>0$), so $A=0$, so $B=0$, so the two functions are linearly independent.
Given any solution $f$ of the differential equation, we can find $A,B$ such that if $g=f-Ae^t-Be^{-2t}$ then $g(0)=g'(0)=0$. But $g$ satisfies the differential equation, so $g$ is identically zero, so $f(t)=Ae^t+Be^{-2t}$.