Question

# Whether T is a linear transformation, that is T : C^1[-1,1] rightarrow R^1 defined by T(f)=f'(0)

Transformation properties
Whether T is a linear transformation, that is $$\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}$$ defined by $$\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}$$

2021-01-29
Approach:
For U and V vector spaces and T is a function from U to V, then T would be considered as linear transformation if for all u, w lies in U and all scalars k as if satisfies the following properties:
$$\displaystyle{T}{\left({u}\ +\ {w}\right)}={T}{\left({u}\right)}\ +\ {T}{\left({w}\right)}$$
$$\displaystyle{T}{\left({k}{u}\right)}={k}{T}{\left({u}\right)}$$
Calculation:
For T is a linear transformation, that is $$\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}$$ defined by $$\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}$$
Such that assume g is also present in the domain of $$\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}$$, that means $$\displaystyle{g}\in{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}$$
Then,
$$\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}$$
$$\displaystyle{T}{\left({g}\right)}={g}'{\left({0}\right)}$$
$$\displaystyle{T}{\left({f}\right)}\ +\ {T}{\left({g}\right)}={f}'{\left({0}\right)}\ +\ {g}'{\left({0}\right)}$$
And,
$$\displaystyle{T}{\left({f}\ +\ {g}\right)}={\left({f}\ +\ {g}\right)}'{\left({0}\right)}$$
$$\displaystyle={f}'{\left({0}\right)}\ +\ {g}'{\left({0}\right)}$$
$$\displaystyle={T}{\left({f}\right)}\ +\ {T}{\left({g}\right)}$$
Thus it shows that $$\displaystyle{T}{\left({f}\ +\ {g}\right)}={T}{\left({f}\right)}\ +\ {T}{\left({g}\right)}$$, that means it satisfies the addition property.
Now, take any scalar be k
And
$$\displaystyle{T}{\left({k}{f}\right)}={\left({k}{f}\right)}'{\left({0}\right)}$$
$$\displaystyle={k}{f}'{\left({0}\right)}$$
$$\displaystyle={k}{\left({f}'{\left({0}\right)}\right)}$$
$$\displaystyle={k}{T}{\left({f}\right)}$$
Thus is also satisfies the scalar property.
So, it satisfies all the condition of the linear transformation.
Hence, the expression of $$\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}$$ defined by $$\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}$$ is linear transformation.
As for $$\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}$$, the property of addition and scalar are satisfied. Thus the expression of $$\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}$$ is linear transformation.