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This shouldn't be too hard, but I'm stuck. Suppose $f$ is a concave function on the interval $[a,b]$, meaning $$\lambda f(x) + (1-\lambda) f(y) \leq f(\lambda x + (1-\lambda) y)$$ for every $x,y \in [a,b]$ and every $\lambda \in [0,1]$. I want to prove that for any $p, q, r \in [a,b]$ with $q \geq r \geq 0$ we have: $$f(p + q) + f(p - q) \leq f(p + r) + f(p - r)$$

This inequality comes up in a paper that I'm reading on random walks, and in that context the function $f$ is piecewise linear. So I'm not willing to assume that $f$ is differentiable, but piecewise smooth is fine if it helps (I don't think it will).

Paul Siegel
  • 9,117

3 Answers3

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Let's calculate first the specific $\lambda$ that "averages" the endpoints $p\pm q$ to give $p\pm r$, i.e. solve for $\lambda$,

\begin{align} \lambda (p+q) + (1-\lambda)(p-q) = \pm r \\ \implies \lambda_{\pm} = \frac {\pm r+q}{2q} \tag{1} \end{align}

Now, using the fact that $q\geq r\geq 0\implies p \pm r\in[p- q,p+q]$,

\begin{align} f(p\pm r) &= f( \lambda_\pm(p+q) + (1-\lambda_\pm)(p-q) )\\ &\geq \lambda_\pm f(p+q) + (1-\lambda_\pm)f(p-q) \\ \end{align}

Summing these and substituting the values in $(1)$,

\begin{align} f(p+r)+f(p-r) &\geq (\lambda_++\lambda_-)f(p+q) + (2-\lambda_+-\lambda_-)f(p-q) \\ &= f(p+q) + f(p-q) \end{align}

Aside

You shouldn't be assuming $p,q,r\in[a,b]$, but rather that $p\pm q\in[a,b]$

davin
  • 3,010
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Since p+q>p+r>p-r>p-q, by concavity, there exists a $\lambda_1, \lambda_2$ such that $$\lambda_1 f(p+q)+(1-\lambda_1)f(p-q)<f(p+r)$$ and $$\lambda_2 f(p+q)+(1-\lambda_2)f(p-q)<f(p-r)$$ and so \begin{equation}(\lambda_2+\lambda_1)f(p+q)+(2-\lambda_1-\lambda_2)f(p-q)<f(p+r)+f(p-r).\end{equation}

Lets now solve for $\lambda_1, \lambda_2$.

We know that $\lambda_1(p+q)+(1-\lambda_1)(p-q)=p+r$ and so $(2\lambda_1-1)q=r.$

Further, $\lambda_2(p+q)+(1-\lambda_2)(p-q)=p-r$ and so $(1-2\lambda_2)q=r.$

Thus we can conclude $\lambda_1+\lambda_2=1$. Inserting this into equation 3, we get

\begin{equation}f(p+q)+f(p-q)<f(p+r)+f(p-r).\end{equation}

Sixter
  • 171
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Oops, this is pretty easy:

Define $\ell(t) = t(p-q) + (1-t)(p+q)$. It is easy to check that: $$\ell \left(\frac{q+r}{2q}\right) = p-r \hspace{1cm} \text{and} \hspace{1cm} \ell \left(\frac{q-r}{2q}\right) = p+r$$ Note that $(q+r)/(2q)$ and $(q-r)/(2q)$ are both in $[0,1]$, and their sum is $1$. It follows that: $$f(p-r) = f \left(\frac{q+r}{2q}(p-q) + \frac{q-r}{2q}(p+q) \right) \geq \frac{q+r}{2q}f(p-q) + \frac{q-r}{2q}f(p+q)$$ and similarly: $$f(p+r) \geq \frac{q-r}{2q}f(p-q) + \frac{q+r}{2q}f(p+q)$$ Adding these two inequalities together gives the result.

Paul Siegel
  • 9,117