Linear functional and bounded norm

Question

Let be given the function (x,y)-->f(x,y) differentiable at (a,b) from R^2

We define the function \begin{array}{l}T:R^2\rightarrow R\end{array}

\begin{array}{l}T\left(h,l\right)=\frac{\partial f\left(a,b\right)}{\partial x}h+\frac{\partial f\left(a,b\right)}{\partial x}l\\ \end{array} Where (h,l)€R^2

How to show if T(h,l) is an linear functional ? Is it bounded? And find the norm ||T||?

I'm trying to work through some example questions from my class but there is no mention of differentiation as an linear transformation anywhere and I'm completely stumped on how to do this.

Any help would be appreciated please.

1. Linearity

\begin{array}{l}T\left(\left(h,l\right)+\left(g,k\right)\right)=\frac{\partial f\left(a,b\right)}{\partial x}\left(h+g\right)+\frac{\partial f\left(a,b\right)}{\partial x}\left(l+k\right)\\ \end{array}

\begin{array}{l}=\frac{\partial f\left(a,b\right)}{\partial x}g+\frac{\partial f\left(a,b\right)}{\partial x}l+\frac{\partial f\left(a,b\right)}{\partial x}g+\frac{\partial f\left(a,b\right)}{\partial x}k\\ \\ \end{array}

\begin{array}{l}=T\left(h,l\right)+T\left(g,k\right)\end{array}

\begin{array}{l}T\left(c\left(h,l\right)\right)=T\left(ch,cl\right)=\\ =\frac{\partial f\left(a,b\right)}{\partial x}\left(ch\right)+\frac{\partial f\left(a,b\right)}{\partial x}\left(cl\right)\\ =cT\left(h,l\right)\end{array}

Answer 1

I think you're letting the derivatives confuse you. As far as you're concerned, $\frac{\partial f(a, b)}{\partial x}$ and $\frac{\partial f(a, b)}{\partial y}$ are just constants.

Your proof is good: the map is linear. To conclude it's a linear functional, simply observe that its codomain is the scalar field $\Bbb{R}$.

Is it bounded? Yes. Every linear map with a finite-dimensional domain is bounded. Note that it can be written as: $$T(h, l) = \left(\frac{\partial f(a, b)}{\partial x},\frac{\partial f(a, b)}{\partial y}\right) \cdot (h, l) = \nabla f(a, b) \cdot (h, l).$$ By Cauchy-Schwarz, $$|T(h, l)| = |\nabla f(a, b) \cdot (h, l)| \le \|\nabla f(a, b)\| \|(h, l)\|$$ with equality if and only if $(h, l)$ and $\nabla f(a, b)$ are non-negative multiples of each other. So, the inequality itself implies that $\|T\| \le \|\nabla f(a, b)\|$, while the extra condition implies that we actually get equality (e.g. take $(h, l) = \nabla f(a, b)$).