Is a decision stump a linear model? Decision stump is a decision tree with only one split. It can also be written as a piecewise function. 
For example, assume $x$ is a vector, and $x_1$ is the first component of $x$, in regression setting, some decision stump can be
$f(x)= \begin{cases} 
 3& x_1\leq 2 \\
 5     & x_1 > 2 \\
   \end{cases}
$
But is it a linear model? where can be written as $f(x)=\beta^T x$? This Question may sound strange, because as mentioned in the answers and comments, if we plot the piecewise function it is not a line. Please see next section for why I am asking this question.

EDIT:


*

*The reason I ask this question is logistic regression is a (generalized) linear model and the decision boundary is a line, also for decision stump. Note, we also have this question: Why is logistic regression a linear model? . On the other hand, it seems not true that decision stump is a linear model.





*

*Another reason I asked this is because of this question: 
In boosting, if the base learner is a linear model, does the final model is just a simple linear model? 
where, if we use a linear model as a base learner, we get nothing more than linear regression. But if we select base learner as a decision stump, we are getting very interesting model. 


Here is one example of decision stump boosting on regression with 2 features and 1 continuous response.

 A: Answers to your questions:


*

*A decision stump is not a linear model.

*The decision boundary can be a line, even if the model is not linear. Logistic regression is an example.

*The boosted model does not have to be the same kind of model as the base learner. If you think about it, your example of boosting, plus the question you linked to, proves that the decision stump is not a linear model.

A: This answer is more verbose than is needed to just answer the question.  I hope to provoke some comments from real experts. 
I once was in a court room and the judge asked (for good reason in context) , if we call a dog's tail a leg, does that mean a dog has 5 legs ?  So what is a linear model ?  
In the context of statistics I've been told by an expert that a linear model means a statistical model constructed from a set of functions $ f_1, f_2, \ldots, f_n$ of the form $ y = \sum a_i f_i $ with the important constraint that the error terms are independent and normally distributed.  With that definition, one can't say if your model is linear because you have given no information about the error term. If one drops the error term constraint, then it is tautologically linear in the function you give or in the function ssdecontrol gives.  However naively, in the context of this question, that may be unsatisfactory.  Any function can be considered as the basis of a linear in that sense.  That is because any space of functions can be turned into a vector space of functions.  
If you are asking on the nose, that is mathematically, if your function linear, then the answer is no.  A linear function is one whose graph is a straight line, while clearly your function doesn't have that property. In answer to the question you pose at the end, that is can one find $\beta$ so that $ f(x) = \beta^{T} x $ , then no.  
Any function of the class you give would satisfy $f(x+y) = f(x) + f(y) $ for any (real) numbers $x$ and $y$. Notice that your function satisfies $ f(1.5) = 3$ and $f(3) = 5$, so $ f(3) \neq f(1.5) + f(1.5)$ as would be required if your function was of the form $f(x) = \beta^T x$.  Notice the class you propose for linear functions is a sub-class of what are usually called linear functions. 
A: No, unless you transform the data.
It is a linear model if you transform $x$ using indicator function:
$$
x' = \mathbb I \left(\{x>2\}\right) = \begin{cases}\begin{align} 0 \quad &x\leq 2\\ 1 \quad &x>2 \end{align}\end{cases}
$$
Then $f(x) = 2x' + 3 = \left(\matrix{3 \\2}\right)^T \left(\matrix{1 \\x'}\right)$
Edit: this was mentioned in the comments but I want to emphasize it here as well. Any function that partitions the data into two pieces can be transformed into a linear model of this form, with an intercept and a single input (an indicator of which "side" of the partition the data point is on). It is important to take note of the difference between a decision function and a decision boundary.
