Proof sketch: For the sake of expressing the idea of the proof, there is no harm in choosing #\rv{a,b}=\rv{0,0}#. In order to ascertain that the stationary point #\rv{0,0}# is a local minimum or local maximum, we check whether this is the case for #f# as a function on every line #l# through #\rv{0,0}#. It is not obvious that this will suffice; it requires more mathematics than we treat here.

We first consider the line through #\rv{0,0}# with equation #x=0#. For the function #f# whose domain is this line, the fact that #\rv{0,0}# is a stationary point means that #f_y(0,0)=0#. But the theory of one variable also has criteria for local minima and maxima:

• If #f_{yy}(0,0)\gt0# then #\rv{0,0}# is a minimum of the restriction of #f# to the line #x=0#
• If #f_{yy}(0,0)\lt0# then #\rv{0,0}# is a maximum of the restriction of #f# to the line #x=0#

For the line #y=0#, the same criteria are valid with respect to the second order derivative with respect to #x#.

The reason that this is not enough for local minimality or maximality of #f# on a two-dimensional open disk around #\rv{0,0}# is that there are many more lines through #\rv{0,0}#: the stationary point #\rv{0,0}# is a local minimum if the double derivative at #\rv{0,0}# of the restriction of #f# to each line through #\rv{0,0}# is positive.

Each line through #\rv{0,0}#distinct from the #y#-axis is given by an equation of the form #y=\lambda x# for a certain number #\lambda#. An arbitrary point on that line has the form #\rv{u,\lambda u}# where #u# is a real number. If we restrict #f# to such a line, then the value of #f# at #\rv{u,\lambda u}# will be equal to #f(u,\lambda u)#. Thus, we can view #x# and #y# as functions of #u# with function rules #x=u# and #y=\lambda u#. The point #\rv{0,0}# is a local minimum of #f# on that line if #f_{uu}(0)\gt0#. The latter expression can be calculated by use of the chain rules for partial differentiation and commuting mixed derviates \[ \begin{array}{rcl}f_{uu}(0)&=&\left.\frac{\dd}{\dd u}\left(f_{x}\cdot x_u+f_{y}\cdot y_u\right)\right|_{u=0}\\&&\phantom{xxxuvw}\color{blue}{\text{chain rule}}\\&=&\left.\left(f_{xx} \cdot x_u+\lambda f_{yx}\cdot x_u + f_{xy}\cdot y_u+\lambda f_{yy}\cdot y_u\right)\right|_{u=0}\\&&\phantom{xxxuvw}\color{blue}{\text{chain rule}}\\&=& \left.\left( f_{xx} +\lambda f_{yx} +\lambda f_{xy}+\lambda^2f_{yy}\right)\right|_{u=0} \\&&\phantom{xxxuvw}\color{blue}{x_u=1\text{ and } y_u=\lambda}\\&=& \left.\left( f_{xx} +2\lambda f_{xy}+\lambda^2f_{yy}\right)\right|_{u=0} \\&&\phantom{xxxuvw}\color{blue}{f_{xy}=f_{yx}}\\&=&f_{xx}(0,0) +2\lambda f_{xy}(0,0)+\lambda^2f_{yy}(0,0) \\&&\phantom{xxxuvw}\color{blue}{u=0 \text{ substituted}}\end {array}\] If this expression is always (that is, for each value of #\lambda#) positive, then #\rv{0,0}# is a local minimum of #f# on any line through the stationary point #\rv{0,0}#. This means that, for each line #l# through the stationary point, a positive number #\varepsilon_l# exists, so #f(x,y)\ge f(0,0)# for all points #\rv{x,y}# on the line #l# at distance at most #\varepsilon# to #\rv{0,0}#. It follows, as mentioned, with slightly more mathematics than we treat here, that there is a positive number #\varepsilon# such that, for each point #\rv{x,y}# in the open disk #S_{\rv{0,0},\varepsilon}# with radius #\varepsilon# and center #\rv{0,0}#, we have #f(x,y)\ge f(0,0)#. But then #\rv{0,0}# is a local minimum of #f#.

So we need to ask ourselves under what condition the expression \(f_{xx}(0,0) +2\lambda f_{xy}(0,0)+\lambda^2f_{yy}(0,0)\) will always be positive (or, for the criterion of a local maximum, always be negative). Because the expression is a quadratic polynomial in #\lambda#, we know that this is the case if the discriminant is negative: \[4f_{xy}(0,0)^2-4f_{xx}(0,0)f_{yy}(0,0)\lt0\tiny.\] After division by #-4#, this reads #H(0,0)\gt0#. This implies the first two statements.

If #H(0,0)\lt0#, then the discriminant is positive, so there are two values of #\lambda# such that the double derivative at #\rv{0,0}# of the restriction of #f# to the line through #\rv{0,0}# is positive for one value of #\lambda# and negative for another. Therefore, the point #\rv{0,0}# is a local minimum of #f# on one line and a local maximum on the other. The conclusion is that #\rv{0,0}# is a saddle point. This completes the sketch of the proof.

If #H(a,b)\gt0# then #f_{xx}(a,b) \cdot f_{yy}(a,b)\gt f_{xy}(a,b) ^2\ge0#. In particular, #f_{xx}(a,b)# and # f_{yy}(a,b)# are nonzero and have the same sign, so #\rv{a,b}# is always a local minimum or local maximum. We conclude that only if #H(a,b)=0#, there is no indication as to the nature of the stationary point #\rv{a,b}#.