Vtyjkyuk

How to prove the real value function $f$ is convex on $[a,d]$ if $f$ is convex on both $[a,c]$ and $[b,d]$? ($a<b<c<d$).

I try to show that: for arbitrary $xin(a,b), zin(c,d)$, and for all $yin (x,z)$ the inequality

$dfracf(y)-f(x)y-xleqslant dfracf(z)-f(y)z-y$

holds. Then $f$ is convex on $[a,d]$.

But it seems to need some inequality tricks, I can't finish it.

This is a fact that is geometrically evident but a pain to prove.

Note that the overlap is essential, if $f(x) = -|x|$, then on the intervals
$[-1,0]$ and $[0,1]$ the function
$f$ is linear hence convex, but clearly $f$ is not convex on $[-1,1]$.

Define
$R(x,y) = f(x)-f(y) over x-y$ for $x neq y$. Note that $R$ is symmetric.
Then $f$ is convex iff $x mapsto R(x,y)$ is monotonically non decreasing for each $y$. (cf. the derivative characterisation of convexity.)

We need to show that for all $t in [0,1]$ that $f((t-1)x+ty) le (t-1)f(x)+t f(y)$.

Suppose $x<y$. If $y in [a,c]$ or $x in [b,d]$ there is nothing to show, so
suppose $x in [a,b), yin (c,d]$.

Choose $m,m'$ such that $b<m'<m<c$. Then
$R(m,x) le R(m,m') le R(y,m') le R(y,m)$. This is where the overlap is used.

Note that
begineqnarray
R(y,x) &=& 1 over y-x [R(m,x)(m-x) + R(y,m)(y-m)] \
&ge& 1 over y-x [R(m,x)(m-x) + R(m,x)(y-m)] \
&ge& R(m,x)
endeqnarray
A similar analysis shows that $R(y,x) le R(y,m)$.

Let $t^* = m-x over y-x$. First suppose $t le t^*$ and let $lambda=t over t^*$.
Note that $x+lambda(m-x) = x+t (y-x)$.
Then
begineqnarray
f(x+t(y-x)) &=& f(x+ lambda(m-x)) \
&le& f(x)+ lambda (f(m)-f(x)) \
&=& f(x)+ t ( y-x over m-x ) ( f(m)-f(x)) \
&=& f(x) + t (y-x) R(m,x) \
&le& f(x) + t (y-x) R(y,x) \
&=& f(x) + t (f(y)-f(x))
endeqnarray
Now suppose $t ge t^*$ and let $mu = t-t^* over 1-t^*$. Note that
$y+(1-mu) (m-y) = x+t(y-x)$. Then
begineqnarray
f(x+t(y-x)) &=& fy+(1-mu) (m-y)) \
&le& f(y)+ (1-mu)(f(m)-f(y)) \
&=& f(y)+ (1-t)( y-x over y-m ) (f(m)-f(y)) \
&=& f(y) + (1-t)(y-x) (-R(y,m)) \
&le& f(y) + (1-t)(y-x) (-R(y,x)) \
&=& f(y) + (1-t) (f(x)-f(y)) \
&=& f(x) + t (f(y)-f(x))
endeqnarray

Using the definition $f(alpha x + (1- alpha) y) le alpha f(x) + (1 - alpha) f(y)$ might be not the simplest approach. Recall, the SOC for convexity, that is, if $f''(x) > 0$ for $x in I$ then inside this interval $I$ function $f(cdot)$ is convex.

Hence, $f''(x) > 0, ~forall x in [a, c]$ and $f''(x) > 0, ~forall x in [b, d] implies f''(x) > 0, ~forall x in [a, d],$ as required.

The points $B= (b,f(b))$ and $C= (c, f(c))$ are the crucial reference points on the graph of $f$.
Let $ell$ be the line connecting $B$ and $C$. Then the two convexity conditions imply that the graph of $f$ restricted to $[a,b]$ is above $ell$, and the graph of $f$ restricted to $[c,d]$ is above $ell$.

But then for any numbers $xin [a,b], yin [c,d]$, the points $X=(x, f(x))$ and $Y=(y, f(y))$ are also above $ell$. Let $P$ be any point on the line segment $XY$. Then the first coordinate of $P$ is in $[a,c]$ or in $[b,d]$: assume that it is in $[a,c]$, the other case is similar.
Then $P$ is above $ell$, as it is above $XC$.

Here I am writing $overlineRS$ to refer to the line segment that passes through $(R, f(R))$ and $(S, f(S))$. And writing $overlineRS(x)$ to refer to the height of the line at $x$ . Saying $overlineRS ge overlineUV$ means $overlineRS(x) ge overlineUV(x)$ within the relevant domain.

We have to esablish $overlineMN$ lies above $f$ for all $M$ and $N$. The cases not covered directly by the convexity assumptions are $M in [A..B)$ and $N in (C .. D]$.

For the sake of contradiction, assume $overlineMN < overlineBN$ (except at point $N$). $M<B$ then implies $f(M) < f(B)$.

Since $C > B$, by convexity assumption within $[B .. D]$, $f(C) le overlineBN(C)$.

But this places $f(B) ge overlineMC(B)$, contradicting convexity within $[A .. C]$. So $overlineMN ge overlineBN$.

By the same argument $overlineMN ge overlineMC$. So $overlineMN(x) ge f(x)$.