Let $f : [0; 1] \to \mathbb{R}$ be a continuous function with $f(0) = f(1).$ Prove that there exists a point $c \in [0, 1/2]$ such that $f(c) = f(c+1/2)$
I know the proof uses the careful application of the Intermediate value theorem to the function $g(x)=f(x)-f(x+1/2)$ establishing a root for the function in $[0,1/2]$. My argument goes as follows:
Claim there exists a $c \in [0,\frac{1}{2}]$ such that $g(c)=0$. Suppose $\nexists$ $c \in \left[0,\frac{1}{2}\right]$ such that $g(c)= 0$, then the continuity of $f$ forces $(g$ is a continuous function$)$ either $g(x)>0$ or $g(x)<0$ for all $x \in \left[0,\frac{1}{2}\right]$. Without loss of generality, assume $g(x)>0$, then $$f(x)>f(x+\frac{1}{2}) \hspace{0.3cm} \forall \hspace{0.2cm} x \in [0,1/2 ]$$ Then setting $x=0$ we get, $$f(0)>f(1/2)$$ and similarly, setting $x=1/2$ we get, $$f(1/2)>f(1)$$ a contradiction to the hypothesis that $f(0)=f(1)$.
Is there any flaw in my argument?
Just simply use the Intermediate Value Theorem from elementary calculus. The one you used is like using a big hammer to pound a little spyder. To this end, again you have: $g(x) = f(x) - f(x+\frac{1}{2})$ is clearly continuous on $[0,\frac{1}{2}]$ since $f$ is. Now $g(0) = f(0) - f(\frac{1}{2}), g(\frac{1}{2}) = f(\frac{1}{2}) - f(1)= f(\frac{1}{2}) - f(0)$. If $g(\frac{1}{2}) = 0 \implies c = \frac{1}{2}$ does it. If not, then $g(0)\cdot g(\frac{1}{2}) = - (f(0)-f(\frac{1}{2}))^2 < 0\implies g(0) < 0, g(\frac{1}{2}) > 0$ or $g(0) > 0, g(\frac{1}{2}) < 0$. Either way use the IVT theorem and there is a $c \in (0,\frac{1}{2})$ such that $g(c) = 0 \implies f(c) = f(c+\frac{1}{2})$.