Let $f[0,3] \to \mathbb{R}$ be increasing such that $f(1)Here is the question.

Let $f[0,3] \to \mathbb{R}$ be monotone increasing such that $f(1)<f(2)$. Is it true that there is $a \in [1,2]$ and $c>0$ such that $f(a+t)-f(a-t) \geq ct$ for all $t \in [0,1]$?

I think this statement is true if $f$ is discontinuous on $[1,2]$. If $f$ has a point of discontinuity on $[1,2]$, it suffices to take the point as $a$. Then $f(a+t)-f(a-t)$ has a positive difference for any $t$ sufficiently small which we can let it be $c$. Then as $t$ increases, $f(a+t)-f(a-t)\ge c \ge ct$ since $t \in [0,1]$.
I just have no idea how to begin for the case where $f$ is continuous in the region $[1,2]$. I'm thinking so far that $f$ is continuous on a compact set $\iff$ $f$ is uniformly continuous. Since $f(1)<f(2)$, there must be some point where the function increases. I'm thinking to choose this point as $a$, but I don't know how to show that $\frac{f(a+t)-f(a-t)}{t} \geq c$. 
Hints appreicated.
 A: Hint:
Suppose $f$ it not only continuos but also absolutely continuos. Then there exists $F \in L^1([1,2])$ such that for all $x,y \in [1,2]$:
$$f(x)-f(y)=\int_y^x F(t)dt$$
Thus, we have:
$$\frac{f(a+t)-f(a-t)}{2t}=\frac{\int_{a-t}^{a+t} F(z)dz}{2t}$$
and so for a.e. $a \in [1,2]$
$$\lim\limits_{t \to 0} \frac{f(a+t)-f(a-t)}{2t}=\lim\limits_{t \to 0}\frac{\int_{a-t}^{a+t} F(z)dz}{2t}=F(a)$$
If it was $F(a)=0$ for all such $a$ then we would have $f$ costant in $[1,2]$ which is not possible. Thus there exists $a \in [1,2]$ with the property we want, i.e. $\lim\limits_{t \to 0} \frac{f(a+t)-f(a-t)}{2t}>0$.
It remains to study the case (more general) when $f$ it is not absolutely continuos.
A: Let $u=\sup\{\,x\mid f(x)\le f(1)\,\}$, $v=\inf\{\,x\mid f(x)\ge f(2)\,\}$. If $u=v$, then $a=u$ and $c=f(2)-f(1)$ works. Hence assume $u<v$.
Then we can replace $f$ with $$\tilde f(x)=f(\tfrac{x-u}{v-u}+1),$$
which is defied (at least) on $[0,3]$, is monotone increasing, and for all $x\in (1,2)$, we have $\tilde f(1)<\tilde f(x)<\tilde f(2)$.  If we can find $\tilde a$ and $\tilde c$ for this $\tilde f$, we can readily turn this into a valid $a$ and $c$ for our original $f$. Concretely, $a=(v-u)(\tilde a-1)+u$ and $c=\min\{\tilde c,\frac{f(2)-f(1)}{2(v-u)}\}$ work, where the min makes sure that this is fine  for $0\le t\le v-u$ as well as for $v-u\le t\le 1$.
Therefore, we assume from now on that $f(1)<f(x)<f(2)$ for all $x\in(1,2)$, in particular for $x=\frac 32$.
Let 
$$ c= \min\left\{\frac {f(2)-f(1)}3, f(2)-f(\tfrac32),f(\tfrac32)-f(1)\right\}>0.$$
Assume this $c$ does not work with any $a\in[1,2]$. Then for each $a\in [1,2]$, we find $t=t(a)\in[0,1]$ with $f(a+t)-f(a-t)<ct$. Clearly, this makes $t(a)>0$. Then the open intervals $(a-t(a),a+t(a))$ cover the compact $[1,2]$, hence there exists a finite subcover $\{\,(u_k,v_k)\mid 1\le k\le n\,\}$. Pick a finite subcover with the minimal number $n$ of intervals.
Note that $$ v_k-u_k\le 2.$$
If $n=1$ and the subcover consists of a single interval $(u_1,v_1)$, then 
$$ f(2)-f(1)\le f(v_1)-f(u_1)<\frac c2(v_1-u_1)<c,$$
contradiction.
If $n=2$, $\frac32$ is in the interval $(u_1,v_1)$ that also covers $1$ or in the interval $(u_2,v_2)$ that also covers $2$.
In the latter case, 
$$f(2)-f(\tfrac32)\le f(v_2)-f(u_2)<\frac c2(v_2-u_2) <c,$$
contradiction.
In the former case similarly,
$$f(\tfrac32)-f(1)<c,$$
contradiction.
Therefore $n>2$.
By minimality, no interval is completely contained in another. In particular, no two intervals have the same left endpoint or the same right endpoint.
Suppose some point $x\in[1,2]$ is covered by at least three of these finitely many intervals, i.e., $x\in (u_1,v_1)\cap (u_2,v_2)\cap (u_3,v_3)$. Pick $i,j$ with $u_i=\min\{u_1,u_2,u_3\}$ and $v_j=\max\{v_1,v_2,v_3\}$. Then  $(u_1,v_1)\cup (u_2,v_2)\cup (u_3,v_3)=(u_i,v_i)\cup (u_j,v_j)$ because $u_j<x<v_i$. Hence a smaller subcover suffices, contradicting minimality. We conclude that each point is in at most two intervals of our subcover. This allows us to index the intervals in such a way that 
$$\begin{align}u_1<u_2<v_1<u_3<v_2&<u_4<\ldots\\
\ldots<u_k<v_{k-1}&<u_{k+1}<v_k<\ldots \\
\ldots &< u_n<v_{n-1}<v_n.\end{align}$$
Note also that
$$ 0\le u_1<1\le u_2,\qquad v_{n-1}\le 2<v_n\le 3$$
as otherwise $(u_1,v_1)$ or $(u_n,v_n)$ would be redundant.
Then for $k>1$,
$$f(v_k)-f(v_{k-1})\le f(v_k)-f(u_k)<\frac c2(v_k-u_k).$$
Summing over $k=2,\ldots,n$,
$$f(v_n)-f(v_1)<\frac c2\sum_{k=2}^n(v_k-u_k). $$
As the $(u_k,v_k)$, $2\le k\le n$, are all $\subset[1,3]$, cover $[1,2]$ at most twice and $[2,3]$ at most once, we conclude
$$f(v_n)-f(v_1)<\frac32c. $$
Similarly, we show
$$f(u_n)-f(u_1)<\frac32c. $$
Since $n>2$, we have $u_n>v_1$ and so
$$\begin{align}f(2)-f(1)&\le f(v_n)-f(u_1)\\&\le f(v_n)-f(u_1)+f(u_n)-f(v_1)\\&<c(v_n+v_{n-1}-u_2-u_1) \\&\le 3c,\end{align}$$
contradiction.
We conclude that our assumption was wrong and there does exists some $a\in[1,2]$ with $f(a+t)-f(a-t)\ge ct$ for all $t\in[0,1]$.
A: Alright, let me post the solution given by the person who set this question.
Split the interval $[1,2]$ into $[1, 1.5], [1.5, 2]$, set $M=f(2)-f(1)$.  Either $f(2)-f(1.5) \geq \frac{M}{2}$ or $f(1.5)-f(1) \geq \frac{M}{2}$. Whichever it is, pick it as your first interval. Say that interval is $[1, 1.5]$. Within this interval, repeat this process using $\frac{M}{4}$. Keeping picking intervals like this, one gets a series of nested intervals, and by Nested Interval Property, there is some point $a$ in the intersection of these intervals.
Given some $t \in \mathbb{R}$, there is some interval $I_n$ sufficiently small enough such that $I_n \subset (a-t, a+t)$. Pick $n$ to be the smallest possible. Then $(a-t, a+t) \subset I_{n-1}$. Since  $I_n \subset (a-t, a+t), f(a+t) - f(a-t) \geq \frac{M}{2^n}$. However, $t \leq \frac{1}{2^{n-2}}$, and selecting $c=\frac{M}{4}$ givesthe desired inequality.
I realise I skipped a lot of steps in the second paragraph, but the key point is the first paragraph.
