# Inequality involving Lipschitz derivative and Taylor's theorem

I have come across an inequality that is supposed to follow from Taylor's theorem and I thought it was obvious until I realized I had the incorrect statement for the multi-variable version written down. How does the statement below follow from Taylors theorem?

Let $$f\in C^1(\mathbb{R}^n)$$ and $$|\nabla f(x) - \nabla f(y)| \leq L |x-y|$$, then by Taylor's theorem $$f(x+h) \leq f(x) + \nabla f(x)\cdot h + \frac{L}{2}\Vert h\Vert^2$$

Though we know the second derivative exists almost everywhere it doesn't seem like we can use it in Taylors theorem. Maybe a bound for the Peano remainder would work?

Any thoughts?

Let $$x, h \in \mathbb{R}^n$$ and set $$g(t) := f(x+th)$$. Then $$f(x + h) - f(x) = g(1) - g(0) = \int_{0}^{1}{dt \, g'(t)} = \int_{0}^{1}{dt \, \nabla f(x + th)\cdot h} \\ = \int_{0}^{1}{dt \,\, \nabla f(x)\cdot h + (\nabla f(x + th) - \nabla f(x))\cdot h} \\ \leq \nabla f(x) \cdot h + \int_{0}^{1}{dt \,\, \Vert(\nabla f(x + th) - \nabla f(x)\Vert \Vert h \Vert} \\ \leq \nabla f(x) \cdot h + \int_{0}^{1}{dt \,\, (L t \Vert h \Vert) \Vert h \Vert} = \nabla f(x) \cdot h + \frac{L \Vert h \Vert^2}{2}.$$