Analysis & Optimization
Problem Set 1
May 24
1. (a) T. This is simply the extreme value theorem. However for those who know some topology, we can
give a more concrete explanation: since [a, b] is compact and f is continuous, f ([a, b]) is compact
and in particular has a minimum (and maximum).1 (To receive credit on this problem it was
sufficient to cite the extreme value theorem.)
(b) F. Consider f (x) = tan x defined on (− π2 , π2 ).
(c) T, since f has a global minimum (due to part a), and a global minimum is automatically a local
minimum.
(d) F. Consider f (x) = −x2 (x − 1)(x + 1) defined on (−2, 2).
(e) F. Consider f (x) = x defined on [0, 1]; here 0 is the global minimum but f 0 (0) = 1.
(f) T. Since I is open and x ∈ I, we can let x + h approach x by taking h either positive or negative.
For h positive,
f (x + h) − f (x)
≥0
h
and for h negative,
f (x + h) − f (x)
≤ 0.
h
Sending h → 0 in both of these inequalities, we have f 0 (x) ≥ 0 and f 0 (x) ≤ 0. So f 0 (x) = 0.
(g) T. This was done in class.
(h) F. Consider f (x) = x3 defined on (−2, 2); f 0 (0) = f 00 (0) = 0 but 0 is not a local minimum.
(i) T. Consider f (x) = sin x defined on (−2π, 2π).
(j) T. Consider f (x) = sin x defined on (−2π, 2π).
(k) F. This is almost tautological; it is impossible that a function would attain two or more minimum
values; one would have to be less than the other.
(l) F. If f has a local minimum at x̂ in the interior of I, then since f 0 is strictly increasing everywhere
in I and f 0 (x̂) = 0, we have f strictly decreasing to the left of x̂ and strictly increasing to the
right of x̂; thus f (x) > f (x̂) for x 6= x̂. We can make a very similar argument2 if x̂ is on the
boundary of I.
(m) T. Consider f (x) = 0 defined on (−2, 2).3
2. (a) By the chain rule,
d
f x, m(x) = fx x, m(x) + fy x, m(x) m0 (x)
dx
1 The
main benefit of this topological explanation is that we only require f to be continuous, as opposed to differentiable.
the argument goes that if I = [a, b] and x̂ = a, then f 0 (a) ≥ 0 and since f 0 is increasing, we have f 0 (x) > 0 ∀x ∈ I,
so that f is increasing throughout the interval and a can be the only minimum. If x̂ = b then f 0 (b) ≤ 0 and we can make the
same type of argument.
3 Note also that unless f is constant, this is false; since f 00 (x) = 0, f (x) must be of the form ax + b, which admits at most
one local minimum on any interval if a 6= 0.
2 Precisely,
1
(b) By the chain rule,
g 0 (t) = fx (th, tk)
d
d
(th) + fy (th, tk) (tk) = hfx (th, tk) + kfy (th, tk)
dt
dt
and
d
d
d
d
(th) + hfyx (th, tk) (tk) + kfxy (th, tk) (th) + kfyy (th, tk) (tk)
dt
dt
dt
dt
= h2 fxx (th, tk) + 2hkfxy (th, tk) + k 2 fyy (th, tk).
g 00 (t) = hfxx (th, tk)
So in particular
g 0 (0) = hfx (0, 0) + kfy (0, 0)
and
g 00 (0) = h2 fxx (0, 0) + 2hkfxy (0, 0) + k 2 fyy (0, 0).
(c) By the chain rule,
d
d
d
(th) + fy (th, tk, t`) (tk) + fz (th, tk, t`) (t`)
dt
dt
dt
= hfx (th, tk, t`) + kfy (th, tk, t`) + `fz (th, tk, t`)
g 0 (t) = fx (th, tk, t`)
and
d
d
d
(th) + hfyx (th, tk, t`) (tk) + hfzx (th, tk, t`) (t`)
dt
dt
dt
d
d
d
+ kfxy (th, tk, t`) (th) + kfyy (th, tk, t`) (tk) + kfzy (th, tk, t`) (t`)
dt
dt
dt
d
d
d
+ `fxz (th, tk, t`) (th) + `fyz (th, tk, t`) (tk) + `fzz (th, tk, t`) (t`)
dt
dt
dt
= h2 fxx (th, tk, t`) + k 2 fyy (th, tk, t`) + `2 fzz (th, tk, t`)
g 00 (t) = hfxx (th, tk, t`)
+ 2hkfxy (th, tk, t`) + 2h`fxz (th, tk, t`) + 2k`fyz (th, tk, t`).
3. (a) The level sets and gradient vectors are drawn below. Fixing x0 ∈ [−5, 5], we have f1 (x0 , y) =
x20 + y 2 , which is clearly minimized at y = 0, with f1 (x0 , 0) = x20 . Since the function f (x) = x2 is
minimized at x = 0, we have that (0, 0) is a local minimum. Note that we could have also seen
this from the computation ∇f1 (x, y) = (2x, 2y), which is only zero for (x, y) = 0; thus the origin
is our only candidate for a local minimum, and it must furthermore be a local minimum since
f1 (0, 0) = 0 and f1 (x, y) > 0 if (x, y) 6= (0, 0).
2
(b) Fixing x0 ∈ [−5, 5], we have f2 (x0 , y) = x20 , which is constant in y; since the function f (x) = x2
is minimized at x = 0, we have that every point (0, y) is a local minimum. We could have also
seen this from the computation ∇f2 (x, y) = (2x, 0), which is only zero for (x, y) = (0, y); thus
these points are our only candidates for local minima, and each must be a local minimum since
f3 (0, y) = 0 and f3 (x, y) > 0 if x 6= 0.
(c) Fixing x0 ∈ [−5, 5], we have f3 (x0 , y) = x20 − y 2 , which is minimized at y = ±5. Since the function
f (x) = x2 − 25 is minimized at x = 0, we have that (0, ±5) are minimum points. Here the
computation of the gradient does not help in finding the minima, since ∇f3 (x, y) = (2x, −2y),
which is only zero at the origin; however f3 (0, 0) = 0 and f3 (x, y) can be made negative by
considering (x, y) arbitrarily close to the origin along the y-axis.
3