# Why must the gradient vector always be directed in an increasing direction?

I don't understand why it is not possible for some function, say for example $f(x,y)$, to not have some point where the partial derivatives combine to a gradient vector that points in a decreasing direction. Why must it always be negated to get a decreasing direction vector?

Intuitively, $f(x + \Delta x) \approx f(x) + \langle \nabla f(x), \Delta x \rangle$. (I'm using the convention that $\nabla f(x)$ is a column vector.) So if $\Delta x = \epsilon \nabla f(x)$ (here $\epsilon > 0$ is tiny), then \begin{align*} f(x + \Delta x) & \approx f(x) + \epsilon \langle \nabla f(x), \nabla f(x) \rangle \\ &= f(x) + \epsilon \| \nabla f(x) \|^2 \\ &\geq f(x). \end{align*}

So when we move a bit in the direction of $\nabla f(x)$, the value of $f$ increases.

I prefer to explain that is slightly different way:

Actually we define gradient to be always pointing to to the maximum increasing direction! take look at the following:

Consider a function $f(x,y)$, then it's full derivative is:

$df(x,y)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)\left(dx,dy\right)=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)\vec{dr}=\left\Vert \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)\right\Vert \left\Vert \vec{dr}\right\Vert \cos\alpha$

so if we consider for simplicity that $\left\Vert \vec{dr}\right\Vert =1$ finaly we get that:

$df(x,y)=\left\Vert \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)\right\Vert \cos\alpha$

So because cosine function is always less or equal to one , we see that the first term is the maximum possible value for our function increase (because that correspond to $\alpha=0$ ) thus if we define this first term as the length of some vector and we name it gradient, then this vector will point out to the direction of maximum possible increase of our function $f(x,y)$.

• Can you please explain this notation on the right ? ∂f∂xdx+∂f∂ydy=(∂f∂x,∂f∂y)(dx,dy) – chris white Oct 27 '12 at 23:30
• I just rewrote the left hand side as a scalar product of two vectors $\overrightarrow{a}\overrightarrow{b}=(a_{x},a_{y})(b_{x},b_{y})=a_{x}b_{x}+a_{y}b_{y}$ – TMS Oct 28 '12 at 10:14

It's not obvious:

Consider the function $$f(x,y):=\cases{0&\bigl((x,y)=(0,0)\bigr),\cr x+y-{4|xy|^{4/3}\over x^2+y^2}&(else) .\cr}$$ Then $f$ is continuous at $(0,0)$ and $\nabla f(0,0)=(1,1)$, but $$f(t,t)-f(0,0)=2t-{4|t|^{8/3}\over 2t^2}=2|t|^{2/3}\bigl(|t|^{1/3}{\rm sgn(t)}-1\bigr)<0\qquad(0<|t|<1)\ .$$ This shows that $f$ is actually decreasing in the direction of the gradient.

Now the considered $f$ is not differentiable at $(0,0)$, and the gradient defined via partial derivatives exists only by coincidence. For any $f$ which is actually differentiable at $(0,0)$ one has $$f(x,y)-f(0,0)=\nabla f(0,0)\cdot (x,y)+o(r)\qquad(r:=\sqrt{x^2+y^2}\to0)\ .$$ Now, if $\nabla f(0,0)=(a,b)\ne(0,0)$ and you choose $(x,y):=(ta,tb)$ with $t>0$ then $$f(ta,tb)-f(0,0)=t(a^2+b^2)+o(t)=t (a^2+b^2)(1+o(1))\qquad(t\to0)\ ;$$ and therefore $f(ta,tb)-f(0,0)$ is $>0$ for sufficiently small $t>0$.

Intuitively:

• if the function is decreasing in one variable, then the partial derivative is negative, so the component vector of the gradient for that variable points in the negative direction - which means increasing function value.

• if the function is increasing in one variable, then the partial derivative is positive, so the component vector of the gradient for that variable points in the positive direction - which means increasing function value.

=> Doesn't matter how the function profile is, the gradient, by definition, points in the increasing direction.

• I don't understand how you started with "the function is decreasing in one variable" and ended up with "which means increasing function value". If you're assuming the function is decreasing in one variable, how is it possible the function value is increasing? – Michael Albanese Jan 15 '16 at 10:09
• f(x,y) is decreasing in x, so the function decreases as you move forward in x. But because the partial derivative with respect to x is negative, the x-component of the gradient points towards origin (move backward in x) - in the direction which makes f to increase. – Cosmin Zamfir Jan 15 '16 at 11:34

Here's another way to look at it: Define $g_v(t) = f(tv)$. Then $g'_v(0)$ represents the rate of increase of $f$ as we walk in the direction of $v$ from $0$. But we can use the chain rule to compute that $g'_v(0) = v \cdot \nabla f(0)$. Looking at the expression $v \cdot \nabla f(0)$, we see that among all unit vectors $v$, the maximum occurs when $v$ points in the same direction as $\nabla f(0)$. We also see that it's at the minimum when it points in the opposite direction. The key to all this is that $f$ be differentiable.

The rate of change of $f$ in a given direction of a unit vector $\vec u$ is given by the directional derivative $D_u f$, where: $D_u f = \left\lVert \nabla f \right\lVert \left\lVert \vec u \right\lVert cos (\theta)$ , for instance, $\theta$ is the angle between $\nabla f$ and $\vec u$. Therefore $D_u f$ is greatest when $cos(\theta)=1$, and this is when $\theta=0$ and $\vec u$ has the same direction as $\nabla f$.

## protected by user99914 Nov 25 '16 at 22:30

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