TITLE: Show that $(C^1)^\perp =\{ 0\}$ in the inner product space of continuous functions
QUESTION [5 upvotes]: In the book of linear algebra by Werner Greub, at page $195,$ it is asked that

Let $C$ be the space of all continuous function in the interval
$0\leq t \leq 1$ with the inner product defined as
$$(f,g) = \int_0^1 f(t) g(t)dt.$$ If $C^1$ denotes the subspace of all continuous
differentiable functions, show that $(C^1)^\perp =\{ 0\}$.

Since I'm dealing with differentiable functions, I thought using integration by parts might help, and I got
$$f(1)G(1)  = f(0)G(0) + \int_0^1 f'(t)G(t) dt, $$
where $G' = g $, but it didn't give me anything useful, as far as I see.
Edit:
This question is asked in abstract linear algebra book, and I have only taken freshman year Calculus courses about functional analysis, so I would appreciate if you give your answer according to this fact.

REPLY [3 votes]: Assume $f\in C,$ with $\int_0^1 f(x)g(x)\, dx = 0$ for all $g\in C^1.$ Define $F(x) = \int_0^x f(t)\,dt.$ Observe $F(1)=0,$ simply because $F(1)=\int_0^1 f(t)\, dt = \int_0^1 f(t)\cdot 1\, dt.$
Let's integrate by parts as you were doing. We get, for all $g\in C^1,$
$$0=\int_0^1 f(t)g(t)\,dt = F(1)g(1)-F(0)g(0) - \int_0^1 F(t)g'(t)\,dt.$$
Because $F(0)=0=F(1),$ we obtain
$$\tag 1 \int_0^1 F(t)g'(t)\,dt=0 \text { for all } g\in C^1.$$
One $g$ we can try in $(1)$ is $g(x) = \int_0^x F(t)\,dt,$ a nice $C^1$ function. Since $g'(x) = F(x)$ by the FTC, we obtain
$$\int_0^1 F(t)^2\,dt = 0.$$
Because $F$ is continuous, the last equality implies $F\equiv 0.$ Therefore $F'(x) = f(x)=0$ for all $x.$