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Chapter 5: Problem 3

Let \(I:=[a, b]\) and let \(f: I \rightarrow \mathbb{R}\) be a continuous function on \(I\) such that for each \(x\) in \(I\) there exists \(y\) in \(I\) such that \(|f(y)| \leq \frac{1}{2}|f(x)|\). Prove there exists a point \(c\) in \(I\) such that \(f(c)=0\).

Short Answer

Expert verified
Hence, there exists a point \(c\) in \(I\) such that \(f(c)=0\).

Step by step solution

01

Begin with function definition

Evaluate the function \(f: I \rightarrow \mathbb{R}\) using the given range, i.e., \(I:=[a, b]\), and property that there exists \(y\) in \(I\) such that |f(y)| \leq \frac{1}{2}|f(x)|.
02

Apply Brouwer's Fixed-Point Theorem

Assume a new function \(g: I \rightarrow I\) that maps every \(x\) in \(I\) to \(y\) which satisfies \(\left| f(y) \right| \leq \frac{1}{2} \left| f(x) \right|\). This function \(g\) is well-defined and continuous. Since \(I\) is a closed interval [a, b], by Brouwer's Fixed-Point Theorem, there exists a point \(c\) in \(I\) such that \(g(c) = c\).
03

Analyze the function at the fixed point

For this fixed point \(c\), \(|f(c)| \leq \frac{1}{2}|f(c)|\), since the function \(g\) maps \(c\) to itself. The only way for this inequality to be satisfied is if \(f(c) = 0\).

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Most popular questions from this chapter

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be continuous on \(\mathbb{R}\) and let \(\beta \in \mathbb{R} .\) Show that if \(x_{0} \in \mathbb{R}\) is such that \(f\left(x_{0}\right)<\beta\) then there exists a \(\delta\) -neighborhood \(U\) of \(x_{0}\) such that \(f(x)<\beta\) for all \(x \in U\).

Let \(K>0\) and let \(f: \mathbb{R} \rightarrow \mathbb{R}\) satisfy the condition \(|f(x)-f(y)| \leq K|x-y|\) for all \(x, y \in \mathbb{R}\). Show that \(f\) is continuous at every point \(c \in \mathbb{R}\).

If \(f:[0,1] \rightarrow \mathbb{R}\) is continuous and has only rational [respectively, irrational] values, must \(f\) be constant? Prove your assertion.

Let \(J:=(a, b)\) and let \(g: J \rightarrow \mathbb{R}\) be a continuous function with the property that for every \(x \in J\), the function \(g\) is bounded on a neighborhood \(V_{\delta}(x)\) of \(x .\) Show by example that \(g\) is not necessarily bounded on \(J\).

Let \(a
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