rudin real analysis chapter3 Q3

Question

Let $\varphi $ be a continuous real function on $(a,b)$ such that $$\varphi(\frac{x+y}{2})\leq \frac{\varphi(x)}{2}+\frac{\varphi(y)}{2} $$ for all $x,y\in (a,b).$ Prove that $\varphi$ is a convex function.

Here is what I know: I know that $\lambda$ have to be arbitrary. And we can use $\lambda$ which its denominator is the power of two and using iteration but I don't know how. I don't know how can I solve the problem for $\lambda=\frac{1}{4}.$ Solving problem for this example can be enough for me too.

Answer 1

For $\lambda = 1/4$,

$$\begin{split} \varphi\left( \frac 14 x + \frac 34 y\right) &= \varphi\left(\frac{1}{2} \left(\frac 12 (x+y)\right) + \frac 12 y\right) \\ &\le \frac 12\left(\varphi \left( \frac 12 (x+y)\right)+ \varphi\left( y\right) \right) \\ &\le \frac 12\left(\frac 12\left( \varphi (x) + \varphi(y)\right) + \varphi (y) \right) \\ &= \frac 14 \varphi (x) + \frac 34 \varphi (y). \end{split}$$

(That is, I first think of $\frac 14 x + \frac 34 y$ as the midpoint of $\frac 12 (x+y)$ and $y$)

Answer 2

$$\frac{1}{4}\varphi(x)+\frac{3}{4}\varphi(y)=\frac{1}{4}\varphi(x)+\frac{1}{4}\varphi(y)+\frac{1}{2}\varphi(y)\geq$$ $$\geq\frac{1}{2}\varphi\left(\frac{x+y}{2}\right)+\frac{1}{2}\varphi(y)\geq\varphi\left(\frac{\frac{x+y}{2}+y}{2}\right)=\varphi\left(\frac{1}{4}x+\frac{3}{4}y\right).$$

Answer 3

We will show by induction that $$\varphi\left(\frac{a}{2^n}x+\left(1-\frac{a}{2^n}\right)y\right)\leq \frac{a}{2^n}\varphi(x)+\left(1-\frac{a}{2^n}\right)\varphi(y)$$ for all integers $0\leq a\leq 2^n$. Note that this statement is given as true for $n = 1$. If the statement is true for $n-1$, then as $$\frac{a}{2^n}x+\left(1-\frac{a}{2^n}\right)y = \frac{1}{2}\left[\left(\frac{\lfloor a/2\rfloor}{2^{n-1}}x+\left(1-\frac{\lfloor a/2\rfloor}{2^{n-1}}\right)y\right)+\left(\frac{\lceil a/2\rceil}{2^{n-1}}x+\left(1-\frac{\lceil a/2\rceil}{2^{n-1}}\right)y\right)\right]$$ considering that $a = \lfloor a/2\rfloor+\lceil a/2\rceil$ for all integers $a$, we have that \begin{align*} \varphi\left(\frac{a}{2^n}x+\left(1-\frac{a}{2^n}\right)y\right) &\leq \frac{1}{2}\varphi\left(\frac{\lfloor a/2\rfloor}{2^{n-1}}x+\left(1-\frac{\lfloor a/2\rfloor}{2^{n-1}}\right)y\right)+\frac{1}{2}\varphi\left(\frac{\lceil a/2\rceil}{2^{n-1}}x+\left(1-\frac{\lceil a/2\rceil}{2^{n-1}}\right)y\right) \\ &\leq \frac{\lfloor a/2\rfloor+\lceil a/2\rceil}{2^n}\varphi(x)+\left(1-\frac{\lfloor a/2\rfloor+\lceil a/2\rceil}{2^n}\right)\varphi(y) \\ &= \frac{a}{2^n}\varphi(x)+\left(1-\frac{a}{2^n}\right)\varphi(y) \end{align*} so we have our induction. Now, consider that for $\lambda\in (0, 1)$, there is a sequence $(a_n)_{n=1}^{\infty}$ such that $0\leq a_n\leq 2^n$ and $\frac{a_n}{2^n}\to \lambda$. Then, $$\frac{a_n}{2^n}x+\left(1-\frac{a_n}{2^n}\right)y\to \lambda x+(1-\lambda)y$$ and $$\frac{a_n}{2^n}\varphi(x)+\left(1-\frac{a_n}{2^n}\right)\varphi(y)\to \lambda\varphi(x)+(1-\lambda)\varphi(y)$$ so as the inequality we proved by induction holds for all $a_n$ and as $\varphi$ is continuous, we have \begin{align*} \varphi(\lambda x+(1-\lambda)y) &= \lim_{n\to \infty} \varphi\left(\frac{a_n}{2^n}x+\left(1-\frac{a_n}{2^n}\right)y\right) \\ &\leq \lim_{n\to \infty} \frac{a_n}{2^n}\varphi(x)+\left(1-\frac{a_n}{2^n}\right)\varphi(y) \\ &= \lambda\varphi(x)+(1-\lambda)\varphi(y) \end{align*} and therefore $\varphi$ is convex.