Is this transformation surjective?

Question

Consider the transformation $T:C_{\mathbb R} [0,1] \to \mathbb R$ defined by $T(f(t)) = \int_0^1 f(t)dt$. Is this transformation surjective? It would be enough to show that $$\mathbb{R} \subseteq\textrm{Ran}(T) = \bigg\{\int_0^1 f(t)\,dt : f(t)\in C_{\mathbb R} [0,1] \bigg\} .$$ Intuitively, this appears to be true since we should always be able to find a function that is large enough such that its integral is the desired real number. I have been trying to think on where to start for the last hours or so, any theorems or hints that might help me get started would be appreciated.

Answer 1

Use the following result:

Let $V$ be a vector space over a field $F$. Any linear transformation (functional) $T:V\to F$ is either zero linear map or surjective.

Proof: Since $T$ is a linear map, $\operatorname{Range}(T)$ is a subspace of $F$. Since the only subspaces of $F$ is either the zero subspace $\{0\}$ or $F$ itself. So the the possibilities for $\operatorname{Range}(T)$ is either $\{0\}$ (and in this case $T=0$) or $\operatorname{Range}(T)=F$ (and in this case $T$ is surjective). This completes the proof.

The above result can be restated as

Any non-zero linear functional $T:V\to F$ is surjective.

For your problem, take $V=(C[0,1],\Bbb R)$ and $F=\Bbb R$ and $T:V\to \Bbb F$ defined by $T(f)=\int\limits_0^1{f(t)dt}$, then clearly $T$ is a non-zero linear map, as $T(f)=1$ with $f(t)=1$, so by the above $T$ is surjective.

Answer 2

Let $c \in \mathbb{R}$. Define $$f_c(t) := c \quad\forall t \in [0,1]$$ Then we have $T(c) = c$, hence T is surjective.

Remark: More generally, every linear functional $T \neq 0$ is surjective. It's not hard to prove this.