I'm having a bit of trouble with a homework question. Here it is:

Let there be a function that $ f:\left[0,1\right]\rightarrow\left[0,1]\right] $ a continuous function. For what values of $\alpha \in \mathbb{R}$ will there be a $c \in \left[0,1\right] $ such that $ f\left(c\right)=\alpha\cdot c$

I just can't fathom how I'm supposed to come up with values for $\alpha $ when the function is not defined for me (as in I don't have an equation for it).

Any hint?


  • $\begingroup$ This is a weird question, or I'm misunderstooding something: it is always true that $\;f(x)=c_xx\;$ for any real function, continuous or not, defined in $\;x\;$ , so I'm guessing this question means $\;\alpha\;$ workds for all $\;c\in [0,1]\;$ ...? So they're asking when a continuous function is a linear one?? Hmmm...weird, as they seem to be wanting a condition on the parameter, not the function itself. $\endgroup$ – DonAntonio Dec 31 '13 at 16:53
  • $\begingroup$ Yes, thats exactly what I don't understand, I'm guessing that they want either to find an arbitrary value for alpha (like 1 for instance) or find a range for which the condition is met (for instance $0<\alpha<1$). Problem is I have no idea what I'm supposed to do if the function isn't fully given to us. $\endgroup$ – user475680 Dec 31 '13 at 16:57
  • $\begingroup$ Is this from some book you can give a reference to? $\endgroup$ – DonAntonio Dec 31 '13 at 16:58
  • $\begingroup$ Nope, not that I know of. Its a homework assignment that as far as I know was not taken from any book $\endgroup$ – user475680 Dec 31 '13 at 17:00
  • $\begingroup$ And is the wording accurate? I'd say it makes no sense since for any $\;\alpha\in\Bbb R\;$ , the function $\;f(x):=\alpha x\;$ obviously fulfills the condition... Something seems to be twisted here... $\endgroup$ – DonAntonio Dec 31 '13 at 17:02

A simple real analysis appoach is sufficient for this.However, I think a more graph-like approach is much better for your develop.
The Approach:
Image $f$ as a curve $y=f(x)$ , that curve is in $Oxy$ and must sasitfy these:
i) $f$ connects 2 point in line $x=0$ and $x=1$ respectively
ii)$f$ lies entirely in the cell $[0;1]\times[0;1]$
iii) (Not necessary for this problem) each line $x=c(0=c=1)$ cuts $f$ at exactly 1 point.

Thus, the question of problem is to find which $\alpha$ so that $y=\alpha x$ and $f$ intersect forall all curve $f$ sastify i) and ii)
. (Try to draw yourself some lines and curves , to know clearlier )

We find that the line $y=\alpha x$ devides the cell in two parts .
Thus, the question of problem occurs if and only if in that cell,line $x=1$ and $x=0$ lies in two different parts( not include endpoints) which are devided by y=\alpha x
(Again, try draw some, it's good for anyone's knowledge)

Which equivalent to $\alpha \ge 1 $ .

In case, you want an alternative proof:
Another proach
For $\alpha <1$ , we choose the function $f_1(x)=1$, thus we easily see that $f_1(x)>\alpha alpha \forall x \in [0;1]$.Hence, these $\alpha$ do not sasity the conclusion.

For $\alpha \ge 1$ , $f(x) \in [0;1]$, therefore $ [ f(0)-\alpha.0][f( 1) -\alpha.1] \le 0$.
Due to the continuousness of $ f(x)-\alpha x$, we imply the conclusion.

  • $\begingroup$ Excellent answer !! +1, I am surprised why this answer was not noticed by many. $\endgroup$ – Paramanand Singh Jan 5 '14 at 10:11

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