Show that $(C^1)^\perp =\{ 0\}$ in the inner product space of continuous functions 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.
 A: 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 $x.$
A: By Stone-Weierstrass we have a sequence $(f_n)$ of polynomials converging uniformly to $f$ on $[0,1]$. We have:
$$\int_0^1 |f(x) - f_n(x)|^2 dx \to 0$$
But:
$$\int_0^1 |f(x) - f_n(x)|^2 dx = \int_0^1 |f(x)|^2dx + \int_0^1 |f_n(x)|^2 dx - 2\int_0^1 f(x) f_n(x) dx = \int_0^1 |f(x)|^2 dx + \int_0^1 |f_n(x)|^2 dx \to 2 \int_0^1 |f(x)|^2dx$$
Therefore $\int_0^1 |f(x)|^2 dx = 0$ and by continuity, $f^2$ is the zero function (i.e. $f$ is the zero function)
A: You can show it by an indirect proof:
Assume there is an $f \in C \cap (C¹)^{\perp}$ with $f(t_+) \neq 0$ for an $t_+ \in [0,1]$.
You can assume that $f(t_+) \gt 0$ (otherwise take $(-f)$ as function):
$$f(t_+)>0$$ 
$f$ is continuous $\Rightarrow$ there is an interval $I$ with a positive length $d>0$ and $t_+ \in I$ such that:
$$f(t) > \epsilon > 0 \mbox{ for all } t \in I$$
Now, choose another smaller interval $I_0 \subset I$ with $t_+ \in I_0$ and length $\frac{d}{2}$ and a function $g \in C¹$ with $g \geq 0$ on $[0,1]$, $g(t) = 1$ on $I_0$ and $g(x) = 0$ for $x \notin I$. So, you get
$$(f,g) = \int_0^1f(t)g(t)\,dt = \int_{I}f(t)g(t)\,dt \geq \int_{I_0}\epsilon \,dt = \epsilon\frac{d}{2} \gt 0$$
This is a contradiction to $f \in (C¹)^{\perp}$.
So, $f(t) = 0$ for all $t \in [0,1]$.
A: You have that $P\subseteq{C^1}$ (where P is the subspace of polinimials defined on $[0,1]$). And this subspace is dense in $C$, so $C^1$ is dense in $C$ too (by stone-weierstrass theorem). with this you can conclude that $(C^1)^{\perp}=\{0\}$.
