Lemma: If $ f $ is as described then $ f(x)=0 \ \forall \ x \in [0, \frac{1}{2c}] $.
Pf: Suppose $ \int_0^\frac{1}{2c} f(t) dt \neq 0 $.
Let $ \epsilon=\frac{c}{2} \int_0^\frac{1}{2c} f(t) dt $
$ f(x) $ is bounded by $ c \| f \|_1 $ so we can let $ M=sup_{x \in [0 \frac{1}{2c}]} f(x). $
Pick $ b \in [0,\frac{1}{2c}] $ such that $ f(b) + \epsilon > M $.
$ \int_0^\frac{1}{2c} f(t) dt \leq \frac{f(b) + \epsilon}{2c} \leq \frac{c}{2c} \int_0^b f(t) dt + \frac{\epsilon}{2c} \leq \frac{1}{2} \int_0^\frac{1}{2c} f(t) dt + \frac{1}{4} \int_0^\frac{1}{2c} f(t) dt $ contradiction
So $ \int_0^\frac{1}{2c} f(t) dt = 0 $.
So $ \forall \ x \in [0 \frac{1}{2c}], f(x) \leq c \int_0^x f(t) dt \leq c\int_0^\frac{1}{2c} f(t) dt = 0 $.
$ \Rightarrow $ Lemma.
So now here's the plan: Take an f that satisfies the hypotheses. If there's some $ \alpha $ such that $ f(\alpha) \neq 0 $. we can shift it over and apply the lemma to the shifted function to get a contradiction.
Here are the details:
