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

Here is the question.

Let $$f[0,3] \to \mathbb{R}$$ be monotone increasing such that $$f(1). Is it true that there is $$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 $$f(1), there must be some point where the function increases. I'm thinking to choose this point as $$a$$, but I don't know how to show that $$\frac{f(a+t)-f(a-t)}{t} \geq c$$.

Hints appreicated.

• Oh right, c must be strictly greater than 0. Yeah monotone increasing in this sense is $a>b \implies f(a) \geq f(b)$, where the inequality isn't strict May 2, 2020 at 10:13

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.

• Sorry, I haven't learnt what is a Lebesgue point is. I'm hoping for a solution someone taking a course in real analysis can understand. May 2, 2020 at 10:39
• @HagenvonEitzen I was wrong, so I edited the answer. Thank you so much for the remark. May 2, 2020 at 12:32

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 $$u. Then we can replace $$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 $$f(1) for all $$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 $$f(a+t)-f(a-t). Clearly, this makes $$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 $$f(2)-f(1)\le f(v_1)-f(u_1)<\frac c2(v_1-u_1) contradiction. If $$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) contradiction. In the former case similarly, $$f(\tfrac32)-f(1) contradiction. Therefore $$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 $$u_j. Hence a smaller subcover suffices, contradicting minimality. We conclude that each point is in at most two intervals of our subcover. This allows us to index the intervals in such a way that \begin{align}u_1 Note also that $$0\le u_1<1\le u_2,\qquad v_{n-1}\le 2 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)\\& 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 $$t\in[0,1]$$.

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.