In one of the questions on my problem sheet, I've been asked to prove that if there is an $f \in C[0,1]$ such that $ \int_0^1 f(t) e^{nt} dt = 0 $ for all $n \in \mathbb{N} $ then $ f \equiv 0 $.
Disclaimer: I have looked at a lot of solutions that discuss a similar problem for example here and all its related links but I still can't figure this one out, even though I've spent quite some time on it. If I've missed something and this question is a duplicate I apologise.
This is what I have worked out so far (which I don't think is very much):
If $f \in C[0,1]$ is such that $ \int_0^1 f(t) e^{nt} dt = 0 $ for all $n \in \mathbb{N} $ then let us consider the partial sum $ S_m(nt) := \sum_{k=0}^m \frac{(nt)^k}{k!}$.
Now we know that this partial sum converges to $e$ uniformly so, for a fixed $ \epsilon > 0 $ there is an $ N \in \mathbb{N} $ such that $\forall m \geq N$ $$\|S_m - e \|_{sup} < \epsilon $$ (I think I've proved it before but if this is incorrect please let me know)
Now that we have this consider:
$$ | \int_0^1 f(t) (S_m(nt) - e^{nt}) dt | \leq \int_0^1 |f(t)| | (S_m(nt) - e^{nt} | \leq \| f \| \| S_m - e \| < \| f \| \epsilon $$
Since epsilon was arbitrary, therefore we have that (I've missed out a few justifications here) $$ \lim_{m \to \infty} \int_0^1 f(t) (S_m(nt) - e^{nt}) dt = 0 $$
and from our assumption, we have that $ \int_0^1 f(t) e^{nt} dt = 0 $ so the above inequality gives us that:
$$ \lim_{m \to \infty} \int_0^1 f(t) S_m(nt) dt = 0 $$
which means that $$ \lim_{m \to \infty} \int_0^1 \sum_{k=0}^m f(t) S_m(nt) dt = \lim_{m \to \infty} \sum_{k=0}^m \int_0^1 f(t)S_m(nt)dt = 0 $$
Now this is where I'm stuck, the last part looks a lot like the elementary application of Weierstrass's Approximation theorem but it's not quite that because in that we assume that $\int_0^1 t^n f(t) \, dt=0, n=1,2,\ldots$ and then prove that $f \equiv 0$ but this is kind of going in the other direction. Am I doing something wrong?
Also I note that if I can somehow prove that $\int_0^1 f^2 (t) dt = 0 $ then we would have that $f \equiv 0$ but I can't seem to figure out any way to get there.
Finally, since this is a problem sheet question I very much would appreciate if you could only give hints rather than the whole answer. Any hints and correction would be much appreciated.
Let $g(t)=f(e)e^{t}$. Then $\int g(t)e^{nt} dt=0$ for all $n \geq 0$. By SWT linear combinations of the functions $e^{nt}: n \geq 0$ are dense in $C[0,1]$. Hence $\int g(t)h(t)dt=0$ for all $h \in C[0,1]$. Hence $g \equiv 0$ and $f \equiv 0$.