Proving that the second derivative of a convex function is nonnegative

Question

My task is as follows:

Let $f:\mathbb{R}\to\mathbb{R}$ be a twice-differentiable function, and let $f$'s second derivative be continuous. Let $f$ be convex with the following definition of convexity: for any $a<b \in \mathbb{R}$: $$f\left(\frac{a+b}{2}\right) \leq \frac{f(a)+f(b)}{2}$$ Prove that $f'' \geq 0$ everywhere.

I've thought of trying to show that there exists a $c$ in every $[a,b] \subset \mathbb{R}$ such that $f''(c) \geq 0$, and then just generalizing that, but I haven't been able to actually do it -- I don't know how to approach this. I'm thinking that I should use the mean-value theorem. I've also thought about picking $a < v < w < b$ and then using the MVT on $[a,v]$ and $[w,b]$ to identify points in these intervals and then to take the second derivative between them, and showing that it's nonnegative.

However I'm really having trouble even formalizing any of these thoughts: I can't even get to a any statements about $f'$. I've looked at a few proofs of similar statements, but they used different definitions of convexity, and I haven't really been able to bend them to my situation.

I'd appreciate any help/hints/sketches of proofs or directions.

Answer 1

Given $f$ is a continuous and using the results from this answer, $f$ can be proven to satisfy: $f(\lambda x_1 + (1-\lambda)x_2) \leq \lambda f(x_1) + (1-\lambda)f(x_2)\ \forall \ \lambda \in [0,1]$

Now, by using Taylor's expansion, $f''(x)$ can be written as: $$ f''(x) = \lim_{h \rightarrow 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2} $$

$f(\frac{1}{2}(x+h) + \frac{1}{2}(x-h)) \leq \frac{1}{2}f(x+h) + \frac{1}{2}f(x-h) \implies 2f(x) \leq f(x+h)+f(x-h)$ or $f(x+h)+f(x-h)-2f(x) \geq 0$.

Since $h^2 \geq 0$ and $f$ being twice differentiable, $f''(x) \geq 0$ follows.

Answer 2

Here's how I ended up solving it. I made reference to this answer for the general principle, and this answer for the part about proving slopes of lines.

Let $m = \frac{a+b}{2}$. Suppose we draw a line (on a two-dimensional Cartesian grid) from $(a,f(a))$ to $(b,f(b))$. Then the point $(m,f(m))$ lies on or below this line because $f$ is midpoint convex: the height of the line at $m$ is given $\frac{f(a)+f(b)}{2}$, which is $ \geq f(m)$ by definition.

Since $(m,f(m))$ lies on or below this line, it follows that the slope of the line from $(a,f(a))$ to $(m,f(m))$ is less than or equal to the slope of the line from $(m,f(m))$ to $(b,f(b))$. We briefly show this:

We want to show that $\frac{f(b)-f(m)}{b-m} \geq \frac{f(m)-f(a)}{m-a}$. Remark that $b-m = \frac{b-a}{2} = m - a$, so the inequality simplifies to $f(b)-f(m) \geq f(m)-f(a)$, which is true since $f(m) \leq \frac{f(m)+f(a)}{2}$.

Having shown that the slope of $((a,f(a)),(m,(f(m))$ is less than or equal to the slope of $((m,f(m)),(b,(f(b))$, we apply the MVT to both of these two lines: $$\exists c \in (a,m) \text{ s.t. } \frac{f(m)-f(a)}{m-a} = f'(c)$$ $$\exists d \in (m,b) \text{ s.t. } \frac{f(b)-f(m)}{b-m} = f'(d)$$ And knowing that the slope of the line joining $((a,f(a)),(m,f(m)))$ is less than or equal to that of the slope of the line joining $((m,f(m)),(b,f(b)))$, we have that $f'(c) \leq f'(d)$. We then apply the MVT once more:

$$\exists z \in (c,d) \text{ s.t. } \frac{f'(d)-f'(c)}{d-c} = f''(z)$$

So $f'(d)-f'(c) = f''(z)(d-c)$. We know that $f'(d) \geq f'(c)$, so $f'(d)-f'(c) \geq 0$. We also know that $d-c > 0$ because $c \in (a,m)$ and $d \in (m,b)$, where $a < b$. So our statement reduces to this inequality: $$f''(z) \geq 0$$ As desired.

Answer 3

I would set up a proof by contradiction. Assuming a single point where $f''(x) < 0$, you can use the continuity of $f''(x)$ to find an interval $[a,b]$, where $f''(x) < 0$ throughout. The intuition is then clear, in the sense that if you draw a concave down segment, then any secant line lies below your curve. I will leave it to you to fill in the details from there.

Answer 4

1.Proof by q-calculus
By using q-calculus,the implication becomes quite straighforward and obvious.
define $d_{q}f(x)=f(qx)-f(x)$,then $f^{'}(x)=lim_{q \to 1}\frac{d_{q}f(x)}{d_{q}x}$ as one may verify that the limit can be deduced by L'hôspital rule.
Now assuming $x_{1}<x_{2}$,without lossing generality,the convex condition has the form $f(kx_{1}+(1-k)x_{2})\leq kf(x_{1})+(1-k)f(x_{2})$.
Subtracting both side with $f(x_{1})$ and dividing both side by $kx_{1}+(1-k)x_{2}-x_{1}$,The inequality now becomes
$\frac{f(kx_{1}+(1-k)x_{2})-f(x_{1})}{kx_{1}+(1-k)x_2-x_{1}}\leq \frac{kf(x_{1})+(1-k)f(x_{2})-f(x_{1})}{kx_{1}+(1-k)x_2-x_{1}}$
now the limiting circumstance
$lim_{k \to 1}\frac{f(kx_{1}+(1-k)x_{2})-f(x_{1})}{kx_{1}+(1-k)x_2-x_{1}}\leq lim_{k \to 1}\frac{kf(x_{1})+(1-k)f(x_{2})-f(x_{1})}{kx_{1}+(1-k)x_2-x_{1}}$
.Both side are the form $\frac{0}{0}$ under limiting situation in which L'hôspital rule is applicable.
Utilizing the L'hôspital rule,one can get
$f^{'}(x_{1}) \leq \frac{f(x_{1})-f(x_{2})}{x_{1}-x_{2}}$ (1)
The same argument applies when the original convex defintion is substracted by $f(x_{2})$,then divided by $kx_{1}+(1-k)x_{2}-x_{2}$ and thence taking limit of $lim_{k \to 0}$,one get the similar inequality but reverse the comparator because $kx_{1}+(1-k)x_{2}-x_{2}\le 0$ when k approaches 0.
$f^{'}(x_{2}) \geq \frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$(2)
With the two inequalities (1) and (2),it's obvious that $\frac{f^{'}(x_2)-f^{'}(x_1)}{x_2-x_1}\geq 0$.Then we get the desire property "for a convex function $f(x)\in C^{2}(X)$ where X might be some compact subset of $R$,$f^{''}(x) \geq 0$ holds in domain where convexity applies."

2.Proof by pure algebraic substitution
The convexity reads like $f(kx_{1}+(1-k)x_{3})\leq kf(x_{1})+(1-k)f(x_{3})$ with $0 \leq k\leq 1$
Let $x_1 \le x_2 \le x_3 $,we intend to select $k=\frac{x_3-x_2}{x_3-x_1}$.
As one may apply this particular k and with some manipulation,one can get
$\frac{f(x_2)-f(x_1)}{x_2-x_1}\leq\frac{f(x_3)-f(x_2)}{x_3-x_2}$
The above inequality obvious implicates $f^{''}(x)\geq 0$.
Here we assume $f(x)\in C^{2}(X)$ where X may be some compact or open subset in R for convexity to hold.

Answer 5

Assuming that $f$ is twice differentiable and $f(\frac{x+y}{2}) \leq \frac{f(x)+f(y)}{2}$, one can show directly that $f'(x)$ is increasing, and hence $f''$ is non-negative. Indeed by induction on $n$ it follows that $f((1-s)x+sy)\leq (1-s)f(x)+sf(y)$ for any $s$ of the form $\frac{k}{2^n}$, $0\leq k \leq 2^n$. In particular, if we let $p_s = x+s(y-x)$ then $$ \tag{$\dagger$} \frac{f(p_s)-f(x)}{p_s-x} = \frac{f(p_s)-f(x)}{s(y-x)}\leq \frac{s(f(y)-f(x))}{s(y-x)} = \frac{f(y)-f(x)}{y-x} $$ and $$ \tag{$\ddagger$} \frac{f(y)-f(p_s)}{y-p_s} = \frac{f(y)-f(p_s)}{(1-s)(y-x)} \geq \frac{(1-s)(f(y)-f(x))}{(1-s)(y-x)} = \frac{f(y)-f(x)}{y-x} $$

Indeed for any $y<z$, let $x_n<y$ be such that $\frac{y-x_n}{z-x_n}=2^{-n}$, that is, $x_n= y-\frac{z-y}{2^n-1}$. Then since $y=x_n+2^{-n}(z-x_n)$ we have $f(y)\leq (1-2^{-n})f(x)+2^{-n}f(z)$ and hence by ($\dagger$) $$ \frac{f(y)-f(x_n)}{y-x_n} \leq \frac{f(z)-f(y)}{z-y} $$ But by the Mean-Value Theorem, $\frac{f(y)-f(x_n)}{y-x_n} = f'(\xi_n)$ for some $\xi_n \in (x_n,y)$, and hence, as $f'$ is differentiable, and thus continuous, since $x_n \to y$ as $n \to \infty$, it follows that $\xi_n \to y$ as $n \to \infty$ and hence $\lim_{n \to \infty} f(\xi_n) = f'(y)$. But now $f'(\xi_n)\leq \frac{f(z)-f(y)}{z-y}$ for all $n$ and thus $f'(y)\leq \frac{f(z)-f(y)}{z-y}$.

But now letting $w_n = 2^{-n}y+(1-2^{-n})z$, we see that since $w_n \to z$ as $n \to \infty$ we have $\lim_{n\to \infty} \frac{f(z)-f(w_n)}{z-w_n} \to f'(z)$ and by $(\ddagger)$ $\frac{f(z)-f(w_n)}{z-w_n} \geq \frac{f(z)-f(y)}{z-y}$ for all $n$, so that $f'(y)\leq \frac{f(z)-f(y)}{z-y}\leq f'(z)$, and hence $f'(x)$ is increasing as required.