Exercise I.A.1

Question

Show that the set of continuous functions
$$\left\{f:\mathbb{R}	o\mathbb{R} | f 	ext{ continuous}ight\}$$
with operations
\begin{align*}
(f+g)(t) &= f(t)+g(t)\
(rf)(t) &= rf(t)
\end{align*}
is a vector space, which we will call $V$.
\begin{enumerate}
\item Let $t_0\in\mathbb{R}$ and let
$$W = \{f\in V | f(t_0) = 0\}$$
Show that $W$ is a subspace of $V$.
\item Let $U=\{f\in V: f(t^2) = f(t)^2 	ext{ for all } t\in\mathbb{R}\}$. Show that $U$ is not a subspace of $V$.
\item Let $X = \{f:\mathbb{R}	o\mathbb{R} | 	ext{$f$ is differentiable}\}$ and show that $X$ is a subspace of $V$.
\end{enumerate}

Toğrul Topal · Accepted Answer

\begin{enumerate}
\item Let $c\in\mathbb{R}$ and $f,g\in W$. We then have $(cf+g)(t_0) = cf(t_0) + g(t_0) = 0$.
\item Consider $\mathrm{id}$. We then have for all $t\in \mathbb{R}: \mathrm{id}(x^2) = x^2 = (\mathrm{id}(x))^2$, yet $\mathrm{id}(cx^2) = cx^2 
eq c^2x^2 =  (\mathrm{id}(cx))^2$ for $c
eq 0,1$. In other words, $U$ is not closed under scalar multiplication.
\item Any linear combination of differentiable functions is a differentiable function (cf. \href{/library/terence_tao/analysis_I}{Analysis I}, Theorem 10.1.13).
\end{enumerate}

Exercise I.A.1

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Exercise I.A.1

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