Let \(f\colon\RR^n\to\RR^n\) be a differentiable function, and let \(L\) be a nonnegative real number. Prove that the following are equivalent:
- For every \(x,y\in\RR^n\),
\[ (f(x)-f(y)).(x-y)\leq L\abs{x-y}^2 \]
- For every \(x,v\in\RR^n\),
\[ Df(x)v.v\leq L\abs{v}^2, \]
where \(Df(x)\) is the derivative of \(f\) at \(x\), and \(.\) denotes the standard inner product of vectors in \(\RR^n\).
Solution
- \(\Longrightarrow\) (ii): Let \(x=y+tv\). Then (i) says
\[ t(f(y+tv)-f(y)).v\leq Lt^2\abs{v}^2. \]
Divide by \(t^2\) and take the limit as \(t\to0\) to deduce \(Df(y)v.v\leq L\abs{v}^2\).
- \(\Longrightarrow\) (i): Let \(\phi(t)=f(y+t(x-y))\) for \(t\in\RR\). Then
\[ \begin{aligned} f(x)-f(y)&=\phi(1)-\phi(0)\\ &=\int_0^1\phi'(t)\,dt\\ &=\int_0^1Df(y+t(x-y))(x-y)\,dt&&\text{(by the chain rule)}, \end{aligned} \]
so
\[ \begin{aligned} (f(x)-f(y)).(x-y)&=\int_0^1Df(y+t(x-y))(x-y).(x-y)\,dt\\ &\leq\int_0^1L\abs{x-y}^2\,dt&&\text{(by (ii))}\\ &=L\abs{x-y}^2. \end{aligned} \]