Subspace of differentiable function

linear algebra

The set of differentiable real-valued functions $f$ on the interval $(0,3)$ such that
$f '(2) = b$ is a subspace of $\mathbb{R}^{(0,3)}$ if and only if $b = 0$. How to prove it?

Best Answer

First suppose that this set is a subspace of $\mathbb{R}^{(0,3)}$, then you know that it is closed under $+$ :

For all $f$ and $g$ in $\mathbb{R}^{(0,3)}$ such that $f'(2)=g'(2)=b$, we have $(f+g) '(2)=b$, and thus: $2b=b$ giving $b=0$.

Now suppose that $b=0$. We need to prove that our set is non-empty and closed under $+$ and $.$ :

Taking any two functions $f$ and $g$ in $\mathbb{R}^{(0,3)}$ and any scalar $k$ in $\mathbb{R}$ such that $f'(2)=g'(2)=b=0$ you need to prove that $(f+g)'(2)=0$ and $(kf)' (2)=0$.

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