Find all values of $\alpha\in\mathbb{R}$ such that for every continuous function $f:[0,1]\to[0,1]$ there exists $c\in[0,1]$ with $f(c)=\alpha\cdot c$

Question

Find all values of $\alpha\in\mathbb{R}$ such that for every continuous function $f:[0,1]\to[0,1]$ there exists $c\in[0,1]$ with $f(c)=\alpha\cdot c$.

pre-attempt

Apparently the solution involves the usage of Intermediate Value Theorem, though I don't understand exactly what's required in this problem. I'd appreciate it if someone could break it down and clarify or give an example.

Answer 1

Hint:

• If $\alpha < 1$, consider $f(x) \equiv 1$.
• If $\alpha \ge 1$, consider the function $g(x) \equiv f(x)-\alpha x$.

Answer 2

I agree with John McClane's answer. $g(x) = f(x) - \alpha x$ is continuous and if $\alpha \geq 1$, $g$ must achieve both a negative (or zero) and a positive value. By IVT, $g(x') =0$ for some $x'$. Then, $f(x') = \alpha x'$.

To understand the idea, draw the graph of $h(c) =\alpha c$ in the interval $[0,1]$ for different values of $\alpha$. Think of what values of $\alpha$ make it inevitable for a continuous $f$ to go through the line $h(c)$.