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
ANSWERED
asked 2021-01-28
Whether T is a linear transformation, that is \(\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\) defined by \(\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}\)

Answers (1)

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,
\(\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}\)
\(\displaystyle{T}{\left({g}\right)}={g}'{\left({0}\right)}\)
Perform additional operation.
\(\displaystyle{T}{\left({f}\right)}\ +\ {T}{\left({g}\right)}={f}'{\left({0}\right)}\ +\ {g}'{\left({0}\right)}\)
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.
Answer:
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.
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