Continuity: f:[a,b] continuous, |f(y)|<=(1/2)|f(x)| for s

BrainMan

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Oct 24, 2007
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I'm confused on this question. Could someone show me what to do? Here it is:

Let f : [a, b] to R be continuous on [a, b]. Suppose that for each x in [a, b], there exists a y in [a, b] such that |f(y)| <= (1/2)|f(x)|. Prove that there exists a point c in [a, b] such that f(c) = 0.

Thanks in advance for any help.
 
Here is an outline. It is rough but you can get the idea.
Suppose that \(\displaystyle \left( {\exists d \in [a,b]} \right)\left[ {f(x) \not= 0} \right]\) otherwise there nothing to prove.
Then \(\displaystyle \left( {\exists d_1 \in [a,b]} \right)\left[ {\left| {f(d_1 )} \right| \le \frac{1}{2}\left| {f(d)} \right|} \right]\).
\(\displaystyle \left( {\exists d_2 \in [a,b]} \right)\left[ {\left| {f(d_2 )} \right| \le \frac{1}{2}\left| {f(d_1 )} \right| \le \frac{1}{{2^2 }}\left| {f(d)} \right|} \right]\).
In general, \(\displaystyle \left( {n > 2} \right)\left( {\exists d_n \in [a,b]} \right)\left[ {\left| {f(d_n )} \right| \le \frac{1}{{2^n }}\left| {f(d)} \right|} \right]\).
Note that \(\displaystyle \left( {\frac{1}{{2^n }}\left| {f(d)} \right|} \right) \to 0\).
If we pick the \(\displaystyle d_n\) with care, we can conclude that \(\displaystyle \left\{ {d_n } \right\}\) has a limit point in \(\displaystyle \left[ {a,b} \right]\).
The from continuity the result follows.
 
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