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Let $(X,d)$ be a compact metric space. Let $f: X \to \mathbb{R}$ be a continuous function such that for each $x \in X$ there is a $y \in X$ with $|f(x)| \leq \frac{1}{3}|f(y)|$. Show that there is a point $c \in X$ such that $f(x)=0$.

I'm stuck on this problem. At first I thought it was an application of the Intermediate Value Theorem. I don't know if that is the case.

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Are you sure it isn't supposed to be $|f(y)|\leq \frac{1}{3}|f(x)|$? Where is the problem from? –  Jonas Meyer Dec 3 '12 at 7:31
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2 Answers

up vote 2 down vote accepted

We know that a continuous image of a compact set is compact, hence your function takes a maximum somewhere, say at $x_0$, that is, $|f(x_0)|\geq |f(x)|$ for all $x\in X$. However, from the hypothesis, we know that $|f(x_0)|\leq\frac{1}{3}|f(y)|$. This implies $3|f(x_0)|\leq |f(y)|$, but since $|f(x_0)|$ is a maximum, $|f(y)|\leq |f(x_0)|$. Combining inequalities gives us $3|f(x_0)|\leq |f(x_0)|$, which implies $2|f(x_0)|\leq 0$. Dividing by 2, and since the absolute value is always nonnegative, we have $|f(x_0)|=0$. Since $|f(x_0)|$ is maximum, we actually have the function is identically zero.

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+1 for proving something stronger. –  Amr Dec 3 '12 at 6:50
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Construct a sequence $x_1,x_2,x_3,...$ such that $f(x_k)\leq(1/3)^{k-1}f(x_1)$. Let $x_{i_1},x_{i_2},...$ be a convergent subsequence. Now we find that $\lim_{k\rightarrow \infty} f(x_{i_k})=0$. Since $f$ is continuous, we find that $f(\lim_{k\rightarrow\infty}x_{i_k})=0$

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This is a natural solution if $|f(y)|\leq \frac{1}{3}|f(x)|$ were written as I initially misread. But rather, it looks like the condition generates a sequence such that $|f(x_k)|\geq 3^{k-1}|f(x_1)|$. (And $f$ must be identically $0$, as Clayton answered and which your argument can be modified to show in an alternative way.) –  Jonas Meyer Dec 3 '12 at 7:34
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