Let $ a,b $ be real numbers such that $a<b$, and $f : [a,b] \rightarrow R $ is continuous such that $ f(a) = f(b) $.
Prove that there exists $\delta > 0$ such that for every $ t \in [0,\delta]$ there is $ x \in [a,b-t] $ that $f(x) = f(x+t)$.
I first defined a new function $g(x) = f(x) - f(x+t)$ for some $t \in [0,\delta]$. And tried to prove that $g(a) = f(a) - f(a+t) < 0 \space, g(b-t) = f(b -t ) -f(b) >0 $ or vice versa. And then use the intermediate value theorem but couldn't proceed any further.
I also tried to take $c$ is the absolute maximum or absolute minimum, and take $ \delta = b - c$ and work around the point $(c,f(c))$ point to find the property.
Thanks for any help.
Let $c,d\in[a,b]$ be such that $f$ has a maximum at $c$ and a minimum at $d$ (these exists because $[a,b]$ is compact). If both $c,d\in\{a,b\}$ then then $f(c)=f(d)$ (because $f(a)=f(b)$), in which case the function is constant and the problem is trivial. Thus assume one of $c$ and $d$ is in $(a,b)$, without loss of generality $c$ (otherwise replace $f$ by $-f$).
Since $c\in (a,b)$ there is a $\delta>0$ such that $[c-\delta,c+\delta]\subset (a,b)$.
For every $t\in [0,\delta]$ define $g_t(x)=f(x+t)-f(x)$. Then $g_t(c-t)\geq 0\geq g_t(c)$ hence there is some $x\in[c-t,c]\subset[a,b-t]$ with the desired property $g_t(x)=0$