If $ c \leq 0 $ there's no work to do, so assume $ c > 0 $ for all that follows.
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) $ and be assured M is finite.
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:
Say f satisfies the hypotheses. [This fixes the constant c] Extend f to $ \mathbb{R} $ by letting $ f(x)=0 $ for $ x<0 $.
Suppose $ \exists \ \alpha $ such that $ f(\alpha) \neq 0 $. Then $ \exists \ a $ such that $ f(a)\neq 0 $ and $ f(x)=0 \ \forall \ x<a-\frac{1}{4c} $.
Let $ h=a-\frac{1}{3c} $
(Note $ h>0 $ by lemma)
Define $ g(x)=f(x+h) $. (Note this defines g on all of $ \mathbb{R} $)
Then g is nonnegative and integrable since f is.
Also, $ g(x)=f(x+h)\leq c\int_0^{x+h} f(t) dt = c\int_{-h}^{x} f(y+h) dy $ (under change of variables $ t=y+h $)
Thus $ g(x)\leq c\int_{-h}^{x} g(y) dy = c\int_{0}^{x} g(y) dy $ (since $ g(x)=0 $ for $ x<0 $)
So if we restrict g to $ [0,\infty) $) we have a function which satisfies the hypotheses of the lemma, thus $ g(x)=0 $ if $ x \in [0,\frac{1}{2c}] $ but if we take $ x=\frac{1}{3c} $ then $ g(x)=f(a)\neq 0 $. contradiction.
So f is identically 0.
--Wardbc 21:56, 21 July 2008 (EDT)
