Suppose $f:R^n \to R$ is differentiable. Let $d\in R^n$ with $∥d∥ = 1,$ show that
$⟨∇f (x), d⟩ ≥ −∥∇f (x)∥.$
For what $d$ does the equality hold?
My solution so far, please let me know if it is correct and let me know if I have potential mistakes.
$⟨∇f (x), d⟩=∇f (x)^Td=\sum_{i=1}^n∇f (x_i)d_i=\frac{df}{dx_1}d_1+\frac{df}{dx_2}d_2+...+\frac{df}{dx_n}d_n$
$∥∇f (x)∥$=$\sqrt{\sum_{i=1}^n∇f (x_i)^2}=\sqrt{(\frac{df}{dx_1})^2+(\frac{df}{dx_2})^2+...+(\frac{df}{dx_n})^2}$
$\frac{df}{dx_1}d_1+\frac{df}{dx_2}d_2+...+\frac{df}{dx_n}d_n\geq-\sqrt{(\frac{df}{dx_1})^2+(\frac{df}{dx_2})^2+...+(\frac{df}{dx_n})^2}$
This equality holds for $d\ge0$
By Cauchy Schwarz, $|\langle\nabla f(x),d\rangle|\leq \|\nabla f(x)\|$. This implies that $-\langle\nabla f(x),d\rangle\leq \|\nabla f(x)\|$ and $\langle \nabla f(x),d\rangle\geq -\|\nabla f(x)\|$.