I apologize beforehand if this is an extremely easy question, I have very limited experience with proving convexity statements. Also, the reason that I choose such a simple example before tackling harder ones is that, if I cannot prove this one, then I have no hope it proving harder convexity statements.

My goal was to shows that the simple $f(x) = x^2$ was convex using the definition only (taking the second derivative and showing is positive is NOT the approach I am looking for).

Recall the definition of a convex function:

$f$ is called convex if: $$\forall x_1, x_2 \in X, \forall t \in [0, 1]: \qquad f(tx_1+(1-t)x_2)\leq t f(x_1)+(1-t)f(x_2).$$

Therefore, I decided to apply both sides of the definition and then compare them:

$$(t x_1 + (1-t) x_2)^2 = t^2x_1^2+2x_1x_2t(1-t)+(1-t)^2x_2^2$$

and I wanted to compare it to:

$$t x_1^2+(1-t)x_2^2$$

I was a little stuck because I didn't know how to show that the inequality is actually true. Intuitively it makes sense because $t \leq 1$ and its square is smaller and thus $t^2x_1^2 \leq t x_1^2$, similarly, $t^2x_2^2 \leq t x_2^2$. However, it was unclear what the cross term did. How do I know that it does not overshoot and make the inequality false?

Thanks for everyones patience.

  • $\begingroup$ $(x-y)^2 \geq 0 \Longrightarrow ... $ $\endgroup$ – Nigel Overmars Nov 17 '14 at 16:11

Subtracting the LHS from the RHS gives $$(t-t^2)x_1^2-2x_1x_2t(1-t)+(t-t^2)x_2^2=t(1-t)(x_1^2-2x_1x_2+x_2^2)=t(1-t)(x_1-x_2)^2\ge 0$$

  • 2
    $\begingroup$ In fact this also shows strict convexity since the inequality is strict if and only if $0 < t < 1$ and $x_1 \ne x_2$. $\endgroup$ – heropup Nov 17 '14 at 16:46

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.