Let f be a convex function on [a,b). Prove that f is bounded from below.

Question

Let $f$ be a convex function on $[a,b)$. Prove that $f$ is bounded from below.

I don't know how to go about it. Applying the definition alone isn't leading anywhere. Can someone drop a hint as to how to begin solving this problem?

This is a problem from Real Analysis, H.L Royden, 3rd edition.

Hint: take two points of the graph of $f$ at abscissas $x_0< x_1$. Show that outside $[x_0, x_1]$, the graph of $f$ is above the line joining these two points. — Gribouillis, Apr 25 '22 at 17:07
Does this answer your question? If $f:I\to \mathbb{R}$ is convex and interval $I$ is bounded, prove that $f$ is bounded below. — 311411, Apr 25 '22 at 18:48

Yuval Peres · Accepted Answer · 2022-04-25T17:30:35.387

4

Let $c=(a+b)/2$ and $d=(c+b)/2$. The definition of convexity implies that in $[a,c]$, the graph of $f$ lies weakly above the straight line $\ell$ connecting $(c,f(c))$ to $(d,f(d))$.

Indeed, if $(x,f(x))$ fell strictly below $\ell$ for some $x \in[a,c]$, then $(c,f(c))$ would be strictly above the line segment connecting $(x,f(x))$ to $(d,f(d))$.

Similarly, in $[c,b]$, the graph of $f$ lies weakly above the line connecting $(a,f(a))$ to $(c,f(c))$.

edited Apr 25 '22 at 17:30

answered Apr 25 '22 at 17:15

Yuval Peres

22,782

OK, I edited my answer. – Yuval Peres Apr 25 '22 at 17:31

Let f be a convex function on [a,b). Prove that f is bounded from below.

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