I am working on exercise 10 of the appendix between chapters 11 and 12 of Spivak's Calculus. The problem is to show that a convex function must be continuous. I would like to check my proof as it is different from the ones I have found so far.

Let $f$ be a function that is convex in $ (a,b)$. Let us assume that $f$ is not continuous in the point $a$.

By the definition of convexity, we have: $ \dfrac{f(x)-f(a)}{x-a}<\dfrac{f(b)-f(a)}{b-a}$

To remove the inequality let $ \dfrac{f(x)-f(a)}{x-a} +h(x) = \dfrac{f(b)-f(a)}{b-a} ~(1)$, where $h(b)=0$ and $h(a)=\dfrac{f(b)-f(a)}{b-a}$

Now let's rearrange equation 1: $ f(x)-f(a) = (x-a)\left(\dfrac{f(b)-f(a)}{b-a} - h(x)\right)$

Taking $lim_{x\to a^+}$ on both sides:

$ lim_{x\to a^+} (f(x)-f(a)) = lim_{x\to a^+} (x-a)\left(\dfrac{f(b)-f(a)}{b-a} - h(x)\right)= 0$

So $~lim_{x\to a^+} f(x)= f(a)$

Thus $f(x)$ is right continuous on $a$

I can use a similar argument to prove that $f(x)$ is left continuous on $b$

Since $f$ is convex in $(a,b)$, it is also convex in $(a+h,b)$ with $h< b-a$. And by letting $h \to b-a$, I prove right continuity over the whole interval.

Similarly, as $f$ in convex in $(a,b)$, it is also convex in $(a,b-k)$ with $k > b-a$. And by letting $k \to b-a$, the whole interval is left continuous.

As any $x_0 \in (a,b)$ can be uniquely expressed as $x_0= a+h = b-k$ and $f$ is right continuous for $a+h$ and left continuous for $b-k$ then $f$ is continuous in $x_0 \in (a,b)$

While writing the question, I have cleaned the logic from what I had initially drafted, so I am more confident about it. Still I am not sure if this logic is correct, as it is longer than any other answer I have found.

  • 1
    $\begingroup$ The proof falls apart at the start. First, you define $h(x) = \frac{f(b) - f(a)}{b-a} - \frac{f(x) - f(a)}{x-a}.$ Then you make a circular argument using $\lim_{x \to a+}(x-a)h(x) = 0$ in proving that $\lim_{x \to a+} (f(x) - f(a))= 0$. $\endgroup$ – RRL Feb 22 '18 at 20:21
  • $\begingroup$ @RRL I knew something was not right. I began with the assumption that $ f(x)\neq f(a)$ so that $h(x)$ is not necessarily defined on $a$. Is the argument circular because $ (x-a)*h(x)$ is zero in $a$ regardless of how $h(x)$ is defined? $\endgroup$ – Amphiaraos Feb 23 '18 at 23:09
  • $\begingroup$ It's circular because $(x-a)h(x) = f(a) - f(x) - (x-a)C$ where $C$ is a constant. You are trying to show that $\lim_{x \to a+} f(x) = f(a)$. You say this follows because $\lim_{x \to a+} (x-a)h(x) = 0$ on the RHS of the equation that follows "Taking $\lim_{x \to a+}$ on both sides ...". So you are using what you are trying to prove. You have not established anything about $h(x)$ yet. $\endgroup$ – RRL Feb 23 '18 at 23:46
  • $\begingroup$ A convex $f:[0,1]\to \Bbb R$ can be discontinuous. E.g. $f(x)=0$ for $x\in [0,1)$ and $f(1)=1$. So you can't separately show that $f$ is left & right continuous. You have to use the fact that the domain of $f$ is open. $\endgroup$ – DanielWainfleet Feb 24 '18 at 2:41

If $f$ is convex on $(a,b)$ then it is relatively easy to show that it is bounded on any closed subinterval.

Let $x$ and $y$ be arbitrary points in $(a,b)$. Assume WLOG $x < y$ and choose $\delta > 0$ such that $a < a+ \delta \leqslant x < y \leqslant b- \delta < b$. Since $f$ is bounded on the closed interval $[a+\delta,b- \delta]$ there exist bounds $m$ and $M$ such that $m \leqslant f(z) \leqslant M$ for all $z$ in the interval.

Defining $z = y + \delta$ and $\lambda = \frac{y-x}{y-x + \delta},$ we have $0 < \lambda < 1$ and $y = \lambda z + (1-\lambda)x$. By convexity

$$f(y) \leqslant \lambda f(z) + (1-\lambda)f(x) = f(x) + \lambda(f(z) - f(x)).$$


$$f(y) - f(x) \leqslant \lambda(f(z) - f(x)) \leqslant \lambda (M - m) = \frac{y-x}{y-x+\delta} (M-m) \leqslant \frac{M-m}{\delta}|y-x|.$$

Switching variable names $x$ and $y$ we get

$$-[f(y) - f(x)] = f(x) - f(y) \leqslant \frac{M-m}{\delta}|x- y| = \frac{M-m}{\delta}|y- x|,$$

and this implies

$$|f(y) - f(x)| \leqslant \frac{M-m}{\delta} |y-x|.$$

Therefore, $f$ is continuous on $(a,b)$ as well as Lipschitz continuous on any closed subinterval.

  • $\begingroup$ Can you please specify more on why $f$ is continuous on $(a,b)$..? $\endgroup$ – Moreblue Nov 10 '18 at 18:10
  • 2
    $\begingroup$ @Moreblue: Sure. Take a fixed $y \in (a,b)$. All $x$ sufficiently close to $y$ lie in some compact interval $[a+\delta,b-\delta]$ where $|f(x) -f(y)| < \frac{M-m}{\delta}|x-y|$ as shown. The parameters $M$,$m$ and $\delta$ are now fixed for all $x$ in the compact interval. If also $|x-y| < \frac{\epsilon \delta}{M-m}$ then $|f(x) - f(y)| < \epsilon$. So $f$ is continuous at each point $y \in (a,b)$. $\endgroup$ – RRL Nov 10 '18 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.