(New page: Lemma: If <math> f </math> is as described then <math> f(x)=0 \ \forall \ x \in [0, \frac{1}{2c}]</math>. Pf: Suppose <math>\int_0^\frac{1}{2c} f(t) dt \neq 0</math>. Let <math>\epsil...) |
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| + | If <math>c \leq 0</math> there's no work to do, so assume <math>c > 0</math> for all that follows. | ||
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Lemma: If <math> f </math> is as described then <math> f(x)=0 \ \forall \ x \in [0, \frac{1}{2c}]</math>. | Lemma: If <math> f </math> is as described then <math> f(x)=0 \ \forall \ x \in [0, \frac{1}{2c}]</math>. | ||
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Let <math>\epsilon=\frac{c}{2} \int_0^\frac{1}{2c} f(t) dt</math> | Let <math>\epsilon=\frac{c}{2} \int_0^\frac{1}{2c} f(t) dt</math> | ||
| − | <math>f(x)</math> is bounded by <math>c \| f \|_1</math> so we can let <math>M=sup_{x \in [0 \frac{1}{2c}]} f(x) | + | <math>f(x)</math> is bounded by <math>c \| f \|_1</math> so we can let <math>M=sup_{x \in [0, \frac{1}{2c}]} f(x)</math> and be assured M is finite. |
Pick <math>b \in [0,\frac{1}{2c}]</math> such that <math>f(b) + \epsilon > M</math>. | Pick <math>b \in [0,\frac{1}{2c}]</math> such that <math>f(b) + \epsilon > M</math>. | ||
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<math>\Rightarrow</math> Lemma. | <math>\Rightarrow</math> Lemma. | ||
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| + | So now here's the plan: Take an f that satisfies the hypotheses. If there's some <math>\alpha</math> such that <math>f(\alpha) \neq 0</math>. | ||
| + | we can shift it over and apply the lemma to the shifted function to get a contradiction. | ||
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| + | Here are the details: | ||
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| + | Say f satisfies the hypotheses. [This fixes the constant c] Extend f to <math>\mathbb{R}</math> by letting <math>f(x)=0</math> for <math>x<0</math>. | ||
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| + | Suppose <math>\exists \ \alpha</math> such that <math>f(\alpha) \neq 0</math>. Then <math>\exists \ a</math> such that <math>f(a)\neq 0</math> and <math>f(x)=0 \ \forall \ x<a-\frac{1}{4c}</math>. | ||
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| + | Let <math>h=a-\frac{1}{3c}</math> | ||
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| + | (Note <math>h>0</math> by lemma) | ||
| + | |||
| + | Define <math>g(x)=f(x+h)</math>. (Note this defines g on all of <math>\mathbb{R}</math>) | ||
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| + | Then g is nonnegative and integrable since f is. | ||
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| + | Also, <math>g(x)=f(x+h)\leq c\int_0^{x+h} f(t) dt = c\int_{-h}^{x} f(y+h) dy </math> (under change of variables <math>t=y+h</math>) | ||
| + | |||
| + | Thus <math>g(x)\leq c\int_{-h}^{x} g(y) dy = c\int_{0}^{x} g(y) dy</math> (since <math>g(x)=0</math> for <math>x<0</math>) | ||
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| + | So if we restrict g to <math>[0,\infty)</math>) we have a function which satisfies the hypotheses of the lemma, thus <math>g(x)=0</math> if <math>x \in [0,\frac{1}{2c}]</math> but if we take <math>x=\frac{1}{3c}</math> then <math>g(x)=f(a)\neq 0</math>. | ||
| + | contradiction. | ||
| + | |||
| + | So f is identically 0. | ||
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| + | |||
| + | --[[User:Wardbc|Wardbc]] 21:56, 21 July 2008 (EDT) | ||
Latest revision as of 09:19, 22 July 2008
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)
