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A function $f : I \rightarrow \mathbb R$ on an interval $I$ is convex if $f((1-t)x+ty)\le (1-t)f(x)+tf(y), \forall x,y \in I, t \in [0,1]$

Assume now that $I$ is an open interval. Show that if $f$ is convex then for each $c \in I$ there exists $m \in \mathbb R$ such that $m(x − c) + f(c) \le f(x)$ for all $x \in I$ , and if in addition $f$ is differentiable at $c$ then $f'(c)$ is the unique $m$ that works. In general, must m be unique?

(we have previously shown that convex functions are continuous)

I think I'm missing the point of the question.

Surely, by the mean value theorem we can find some $y\in (x,c)$ such that $f'(y)=\frac{f(x)-f(c)}{x-c}$ setting $m=f'(y)$ we get $m=\frac{f(x)-f(c)}{x-c}, m(x-c)+f(c)=f(x)$ such an $m$ fits the inequality we're asked to show.

I also cannot see why $f'(c)$ would be the unique solution or why we would ever get a unique solution of the inequality at all for that matter.

Clearly I'm misunderstanding the question, so if someone could point out where I would appreciate that.

Thank you

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    $\begingroup$ Have you tried drawing a picture? Try drawing a line $m(x-c)+f(c)$ (i.e. a line through $(c,f(c))$) that is never above the graph of $f$ on the interval $I$. If $f$ is differentiable at $c$ then there is only one such line: the tangent line (i.e. the line with $m=f'(c)$). $\endgroup$ – smcc Aug 13 '16 at 21:54
  • $\begingroup$ I should have said: "if $f$ is differentiable and convex then there is only one such line". You did not use convexity yet in your attempt at answering the question. $\endgroup$ – smcc Aug 13 '16 at 22:03
  • $\begingroup$ Actually now that I think about it more, what's the significance in it being convex? $\endgroup$ – Aka_aka_aka_ak Aug 13 '16 at 22:07
  • $\begingroup$ The question is really getting you to show that for differentiable functions convexity can be characterized as the property that the graph of the function never lies below its tangent lines. $\endgroup$ – smcc Aug 13 '16 at 22:10
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As smcc pointed out in their comment, we can think of $y=m(x-c)+f(c)$ as a line passing through the point $\left(c, f(c)\right)$ with slope $m$. Let's assume that this line has a second intersection with the curve $f(x)$: $\left(b, f(b)\right)$. Then, we can rewrite the equation of the line using the fact that $\displaystyle m=\frac{f(b)-f(c)}{b-c}$: $$y=\frac{f(b)-f(c)}{b-c}(x-c)+f(c)$$ We can use the substitution $\displaystyle t=\frac{x-c}{b-c}$ to get the following equation:

$$y=tf(b)+(1-t)f(c)$$

Now, we use the convex condition which was mentioned in the OP: $$f(tb+(1-t)c)\leq tf(b)+(1-t)f(c) \text{ for }t\in[0,1] $$

Unless $f(x)$ is the equation of a line, equality is only reached when $t=0$ or $t=1$. Ergo, for all $x$ between $b$ and $c$, $m(x-c)+f(c)>f(x)$. The only case in which this does not happen is when the line $y=m(x-c)+f(c)$ has no second intersection. In other words, $m=f'(c)$.

In the case that $f(x)$ is a line, the question should be easy to answer.

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  • $\begingroup$ @user307463 Although I appreciate that you've accepted my answer, I would like to point out that I never proved that such an $m$ exists for general convex functions. You should read through A.G.'s answer for a proof of existence. $\endgroup$ – Hrhm Aug 14 '16 at 22:41
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Here is an idea how you can prove it.

  1. Take a point $a\in I$ to the left of $c$ and a point $b\in I$ to the right, i.e. $a<c<b$, and draw two secant lines: $\ell_a$ through $(a,f(a))$ and $(c,f(c))$ and $\ell_b$ through $(c,f(c))$ and $(b,f(b))$. Denote $k_a$ and $k_b$ their slopes respectively.
  2. By convexity, the graph of $f$ is under $\ell_a$ on $[a,c]$ and above it outside $[a,c]$. Similarly, the graph of $f$ is under $\ell_b$ on $[c,b]$ and above it outside $[c,b]$. It gives, in particular, that $k_a\le k_b$.
  3. Let $a\to c^-$. The slope $k_a$ is monotonically increasing (again by convexity) and bounded from above by $k_b$, hence, converges to some value $M_1$. Similar argument when $b\to c^+$ gives that $k_b\to M_2$.
  4. Since $k_a\le k_b$ we get $M_1\le M_2$. Now any $m\in[M_1,M_2]$ works.
  5. If $f$ is differentiable at $c$ then $M_1=M_2$.

P.S. The interval $[M_1,M_2]$ is called subdifferential of $f$ at $c$.


In your argument, the point $y\in (x,c)$ you found by the mean value theorem depends on $x$, hence $m=f'(y)$ also does. However, the value of $m$ in the claim must be the same for all $x\in I$.

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