Why is a gradient not a linear functional? I'm reading this book and on page 104 they define:


Afterwards they said that if $m=1$, the function $f$ is real-valued and $T$ is the gradient which has to be a linear functional according to the definition.
The problem is this is not true in general, see for example:
$f:\mathbb R^2\to \mathbb R$ defined as $f(x,y)=\sin x$. The gradient is $\nabla f=(\cos x,0)$ which is not a linear functional.
 A: The gradient of $f$ is not lienar, but the gradient operator is linear.
i.e., $\nabla(f+g)=\nabla f+\nabla g$ and $\nabla cf =c\nabla f$
A: Technically, the gradient at each point should be a linear functional for just this reason. What that means is that, if you think about it from a linear algebra perspective, is that the gradient is a covector, or dual vector, or row vector, i.e. it inhabits $(\mathbb{R}^n)^{*}$, because such a linear functional is the definition of a covector.
You can "check" this by thinking about how the gradient transforms when the input to $f$ is subject to a coordinate transformation. Namely, because of the chain rule, the gradient's components transform covariantly, instead of contravariantly; thus it must be a covector. In a sense, your elementary Calc text lied to you when it called this a vector (but not entirely - covectors are vectors, they just live in a different vector space than the things that are commonly called "vectors" at that stage. That's the "lie". However, the two spaces are isomorphic, so you can "ignore" this in a sense ... until you can't, as here.).
