If $(f_n)\to0$ in $L^2$ and the derivatives converge in $L^2$, must the limit be 0? Question

Let $f_n:[0,1]\to\mathbb R$ be infinitely differentiable functions such that
$$\lim_{n\to\infty}\int_0^1\!f_n^2\,dx=0,$$
and for all $\varepsilon>0$ there exists $N\in\mathbb N$ such that for all $n,m\geq N$,
$$\int_0^1\!(f'_n-f'_m)^2\,dx<\varepsilon.$$
Must it be true that
$$\lim_{n\to\infty}\int_0^1\!{f'_n}^2\,dx=0?$$

Motivation
The following is an exercise for a first course in functional analysis: Let $\Omega\in\mathbb R^n$ be compact. Prove that for all $f\in C(\Omega)$ there exists $u\in H^1(\Omega)$ such that
$$\int_\Omega\!\nabla u\cdot\nabla\phi\,dx=\int_\Omega\!f\phi\,dx\quad\text{for all }\phi\in C^\infty(\Omega),$$
where $H^1$ is the completion of $C^\infty$ under the inner product $\langle f,g\rangle=\int_\Omega(fg+\nabla f\cdot\nabla g)\,dx$.
I can solve the question provided that $[f,g]=\int_\Omega fg\,dx$ in $C^\infty$ extends to an inner product on $H^1$. If $(f_n),(g_n)$ are sequences in $C^\infty$ which are Cauchy (with respect to $\langle\cdot,\cdot\rangle$), it is not hard to show that $\lim_{n\to\infty}[f_n,g_n]$ exists and is independent of choice of representative Cauchy sequences, so $[\cdot,\cdot]$ is well-defined in $H^1$.
Most of the axioms of an inner product space for $(H^1,[\cdot,\cdot])$ can be recovered from those in $(C^\infty,[\cdot,\cdot])$, except $[f,f]=0\implies f=0$. This is equivalent to the following statement about Cauchy sequences:

Let $f_n\in C^\infty(\Omega)$ such that $[f_n,f_n]=\int_\Omega\!f_n^2\to0$, and $\forall\varepsilon>0\,\exists N\in\mathbb N\,\forall n,m\geq N$
$$\langle f_n-f_m,f_n-f_m\rangle\to0,\quad\text{ie. }\int_\Omega\!\nabla(f_n-f_m)\cdot\nabla(f_n-f_m)\,dx\to0.$$
Show that $\langle f_n,f_n\rangle\to0$, ie. $\int_\Omega\nabla f_n\cdot\nabla f_n\,dx\to0$.

Taking the simplest case, $\Omega=[0,1]$, gives the question stated above.
Edit

I am looking for solutions with preferably minimal use of tools from functional analysis. In particular, the concept of weak convergence is only covered later in the course, and weak derivatives are not covered at all, so I suppose we are not expected to be familiar with them for this question.
Right now it appears to me that the question is about a sequence of functions in $C^\infty$, or even $L^2$, so there should be a direct solution using tools from real analysis or measure theory. If this is incorrect, it would be very helpful if someone can answer the following (admittedly much more vague) question instead:
Question': why does the above question "live" in $H^1$, instead of $C^\infty$ or $L^2$?

 A: Due to your assumptions $(f_n)$ converges in $L^2(\Omega)$, and $(f_n')$ is a Cauchy sequence in $L^2(\Omega)$ hence convergent. That is, $f_n \to f$ and $f_n'\to g$ in $L^2(\Omega)$.
Let $\phi\in C_c^\infty(\Omega)$ be given. Passing in the equation
$$
\int_\Omega f_n \phi' = -\int_\Omega f_n' \phi
$$
to the limit yields $f'=g$ in the sense of weak derivatives. Then it is not hard to conclude $f_n\to f$ in $H^1(\Omega)$.
[In the question $f=0$, then also $f'=g=0$ follows.]
A: We are given $f_n \to 0$ in $L^2 = L^2[0,1].$ We are also given $(f_n')$ is Cauchy in $L^2.$ Since $L^2$ is complete, $f_n'$ converges in $L^2$ to some $g\in L^2.$ (Let's note for later use that this implies $f_n' \to g$ in $L^1.$)
You are asking if $g=0$ in $L^2.$ Suppose this fails. Then $g\ne 0$ on some set of positive measure. WLOG, $g>0$ on some $E\subset [0,1]$ with $m(E)>0.$
Apply the Lebesgue differentiation theorem to $g$ on $E:$ For a.e. $a\in E,$ we have
$$\tag 1 \lim_{h\to 0^+} \frac{1}{h}\int_a^{a+h} g(x)\,dx = g(a)>0.$$
Now $f_n \to 0$ in $L^2$ implies there is a subsequence $f_{n_k}$ that converges pointwise to $0$ a.e. It follows that there exists $a_0\in E$ such that i) $(1)$ holds with $a=a_0$ and ii) $f_{n_k}(a_0) \to 0.$
For this $a_0$ there then exists $\delta > 0$ such that
$$\tag 2 \int_{a_0}^{a_0+h} g(x)\,dx > (g(a_0)/2)h\,\,\text { for } 0<h<\delta.$$
Again using $f_{n_k} \to 0$ a.e., we have $f_{n_k}(a_0+h_0) \to 0$ for some $h_0\in (0,\delta).$
We have arrived at two incompatible results: By $(2)$ we have
$$\tag 3 \int_{a_0}^{a_0+h_0}f_{n_k}'(x) \,dx > (g(a_0))/2)h_0$$
for large $k,$ but by smoothness, $(3)$ equals $f_{n_k}(a_0+h_0)-f_{n_k}(a_0),$ and both of these terms $\to 0$ as $k\to \infty.$
This contradiction shows $g=0$ in $L^2,$ and we're done.
A: While this question does start to hint at the utility of Sobolev spaces, you do not really require that machinery here. First, recall that convergence in $L^2$ implies convergence pointwise almost everywhere along a subsequence. So there exists $(n_k)$ such that $f_{n_k}(x)\to0$ for all $x\in A$, where $A\subset[0,1]$ has full measure.
Next, we know that $(f_n')$ is a Cauchy sequence in $L^2(0,1)$, so $f_n'\to g$ in $L^2$ for some $g\in L^2(0,1)$. For any $a,b\in(0,1)$, we have
$$\left|\int_a^bf_n'(t)dt-\int_a^bg(t)dt\right|\le\int_0^1|f_n'-g|dt\le\left(\int_a^1|f_n'-g|^2dt\right)^{1/2}\to0$$
as $n\to\infty$. Since $f_n(b)-f_n(a)=\int_a^bf_n'(t)dt$, choosing $a,b\in A$ and $n=n_k$ implies
$$\left|\int_a^bg(t)dt\right|\le|f_{n_k}(b)-f_{n_k}(a)|+\left|\int_a^bf_n'(t)dt-\int_a^bg(t)dt\right|\to0$$
as $k\to\infty$. From here it is straightforward to verify that $\int_Bg(t)dt=0$ for any Borel $B\subset[0,1]$, and so $g=0$ almost everywhere. This completes the proof.
