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There is one I could not understand in the process of proving the first order convexity condition, which is $f$ is convexif and only if dom $f$ is convex and $f(y) \gt f(x) + \nabla f(x)^T (y-x)$ holds for all $x, y\in$ dom $f$.

It is $$g'(t) = \nabla f(ty + (1-t)x)^T (y-x),$$ where $g(t) = f(ty + (1-t)x)$, and $x,y$ are in $\Bbb{R}^n$, but $t$ is in $\Bbb{R}$.

$t$ is not a vector but a scalar value!!

Why does there exists $\nabla$ in the equation of $g'(t)$??

Danny_Kim
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  • $\nabla$ differentiates with respect to x and y, not t. t will be treated as a constant. f(ty+ (1-t)x) – user247327 Apr 19 '17 at 12:33
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    Because $f$ is a function of several variables and the chain-rule says that $\frac{d}{dt}f({\bf z}) = (\nabla f({\bf z})) \cdot \frac{d{\bf z}}{dt}$. Note that $f$ is a scalar so the gradient is a vector and the dot-product on the right makes this whole term a scalar which agrees with the left hand side. – Winther Apr 19 '17 at 12:34

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