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let $\Omega$ be an open subset of $\mathbb{R}^n$ and $f$ a smooth convexe function $f:\Omega\rightarrow \mathbb{R}$, and let $x_0\in\Omega$ we have always this inequality :

$$\forall x\in\Omega\space :f(x)\geq f(x_0)+<\nabla f , x-x_0>$$

i want to know what is the gerometric interpretation of the right side of this inequality (when n=1 it is the tangent line normally under the graph , but when n>1 is it the equation of the tangent plane or what )

thanks ...

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You are correct. If $f \colon \Omega \to \mathbf R$ is differentiable at $x_0$, then $$ T(x) := f(x_0) + Df(x_0)(x-x_0) = \langle\nabla f(x_0), x-x_0\rangle + f(x_0) $$ is the tangent hyperplane to $f$ at $x_0$. Note that $f$ and $T$ equal at $x_0$ and their first derivatives also do, as expected. We have, as in the $n=1$ case, that $$ f(x_0) = T(x_0), \quad Df(x_0) = DT(x_0). $$

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  • $\begingroup$ thanks for unswering , but i see that here $Df(x_0)(x-x_0) = \langle\nabla f(x_0), x-x_0\rangle$ then what doese $Df(x_0)(x-x_0)$ geometriclly mean, and also $\nabla f(x_0)$ (for $\nabla f(x_0)$ i think it s the normal vector of the tangent plane in the point $(x_0,f(x_0)$) ? $\endgroup$ – M.luffy Feb 15 '16 at 10:14
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$$<\nabla f , x-x_0>$$ is indeed an expression that evaluates to $0$ when the vector $x-x_0$ is orthogonal to the gradient $\nabla f$, i.e. when $x$ lies in the plane orthogonal to the surface normal at $x_0$, which is the tangent plane.

So the formula can be interpreted as the surface of "altitude" $z=f(x)$ lies "above" its tangent plane at $x_0$.

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