1.2.9 · D5Calculus & Optimization Basics
Question bank — Local vs global minima - maxima
This bank hunts the sneaky ideas around local vs global extrema. Every item is a one-line reveal: read the prompt, commit to an answer in your head, then click. The answers give reasons, not verdicts — because "true" or "false" is worthless without the why.
Before we start, a tiny shared vocabulary so no symbol sneaks in undefined:
True or false — justify
TF1. Every global minimum is also a local minimum.
True — the global floor is lower than everything, so it is certainly lower than its immediate neighbors, which is exactly the local condition.
TF2. Every local minimum is also a global minimum.
False — a local min only beats its neighborhood; a deeper valley elsewhere can undercut it. Global is the stronger claim.
TF3. If then is a minimum.
False — flat slope only means critical; it could be a maximum, a saddle, or an inflection like at the origin.
TF4. A function can have two different points that are both global minima.
True — global minima must share the same lowest value, but that value can be hit at several places (e.g. reaches at both and ).
TF5. On a closed interval, the global maximum must occur at a critical point.
False — it can also sit at a boundary/endpoint, where the slope need not be zero at all.
TF6. A positive-definite Hessian at a critical point guarantees a global minimum.
False — it guarantees a local min (bowl-up in every direction), but says nothing about deeper valleys far away.
TF7. In very high dimensions, saddle points tend to outnumber local minima.
True — for a random critical point, all eigenvalues being the same sign gets exponentially unlikely as dimension grows, so mixed-sign (saddle) points dominate.
TF8. A convex loss surface can still trap Gradient Descent in a bad local minimum.
False — convex means one connected bowl; every local minimum is the global minimum, so there is nothing bad to get trapped in.
TF9. Reaching a local minimum in deep learning always means the model is under-performing.
False — in overparameterized networks many local minima are nearly as good as the global one; a 95.0% vs 95.2% gap is often irrelevant.
TF10. If at a critical point, the point cannot be a minimum.
False — the second-derivative test is merely inconclusive there; has yet is a genuine minimum.
Spot the error
SE1. "I set the gradient to zero and found , so training is done — that's the best weights."
Zero gradient only means critical; in a neural loss it is very likely a saddle, not the best point. You must check curvature (Hessian) or, in practice, whether loss is still improving.
SE2. "The Hessian has a negative eigenvalue, so this critical point is a maximum."
One negative eigenvalue only rules out a minimum. A maximum needs all eigenvalues negative; mixed signs make it a saddle, not a max.
SE3. " is a critical point of , therefore it is a local extremum."
It is an inflection point — the function keeps rising through it, going down then up in slope but never turning. Critical does not imply extremum.
SE4. "Since gradient descent only steps downhill, it can never increase the loss, so it must reach the global minimum."
Downhill-only is exactly why it can get stuck — it settles in the first valley it enters and cannot climb out to reach a deeper one. See Momentum and Nesterov for escape tricks.
SE5. "The loss surface plot shows one deep valley, so there's only one minimum."
A 2-D or 3-D plot is a projection/slice of a million-dimensional surface; it hides most of the geometry. You cannot read the true number of minima off a low-dimensional picture.
SE6. "The learning rate doesn't affect which minimum we land in, only how fast we get there."
A large learning rate can jump over narrow valleys and land in an entirely different basin, so it changes the destination, not just the speed.
SE7. "Positive definite means every entry of the Hessian is positive."
No — positive definite is about eigenvalues (equivalently for all ). A Hessian with negative entries can still be positive definite.
Why questions
WHY1. Why is a necessary condition for a smooth local minimum?
If the gradient were nonzero, stepping in direction would strictly lower , contradicting that no nearby point is lower.
WHY2. Why do we need the second-order (Hessian) test after finding critical points?
The gradient tells us where the ground is flat but not how it curves; the second-order term decides bowl-up (min), bowl-down (max), or mixed (saddle).
WHY3. Why is there no purely local test for global optimality?
A neighborhood only sees nearby points; a deeper valley an arbitrary distance away is invisible to any local check, so you must compare all critical points and boundaries.
WHY4. Why does non-convexity make neural network training genuinely hard?
Non-convex means countless critical points; you cannot guarantee the one you reach is the global optimum, and exhaustive search over millions of weights is impossible.
WHY5. Why does momentum help escape poor local minima?
Accumulated velocity lets the optimizer roll through shallow dips and small barriers instead of halting the instant the slope flattens, much like a ball with inertia.
WHY6. Why can two identical network architectures reach different final accuracies?
Random initialization drops them into different basins, and the non-convex landscape means each basin can converge to a different local minimum with a different loss.
WHY7. Why do adaptive methods like the Adam Optimizer matter near saddle points?
Near saddles the gradient is tiny, so plain descent crawls; per-parameter scaling amplifies movement along flat directions, helping escape faster.
WHY8. Why does generalization not simply mean "reach the global minimum of training loss"?
The global training minimum can overfit; a slightly higher-loss ("flatter") minimum often generalizes better to unseen data. Lowest training loss ≠ best model.
Edge cases
EC1. What happens on a perfectly flat function ?
Every point is simultaneously a global minimum and a global maximum, and the gradient is zero everywhere — no unique extremum exists.
EC2. Where can extrema of a function on a closed bounded domain live besides critical points?
On the boundary, where the slope need not vanish; you must always evaluate endpoints/edges and compare them against interior critical values.
EC3. A critical point where the Hessian has a zero eigenvalue (semidefinite) — what do we conclude?
The test is inconclusive along that flat direction; the point could be a min, a saddle, or a degenerate valley — you need higher-order terms to decide.
EC4. A "valley" that is completely flat along one direction (a ravine floor) — is the minimum unique?
No — a whole line of points shares the lowest value, so the global minimum is a set, not a single point; this is common with symmetric or redundant parameters.
EC5. What is the extremum status of for , where the derivative doesn't exist?
It is a genuine global minimum, even though is undefined — the smooth-gradient test simply doesn't apply, so definitions (not derivatives) must decide.
EC6. If the domain is unbounded and (a straight ramp), where is the global minimum?
There is none — decreases forever as , so no lowest value exists; global extrema can fail to exist entirely.
EC7. Does adding regularization change the extrema of the loss?
Yes — it reshapes the landscape (e.g. pulling weights toward zero), moving critical points and often making the effective surface better-behaved and easier to optimize.
Recall Fast self-check
Global implies local, but never the reverse? ::: Correct — global is the stronger condition. Zero gradient guarantees a minimum? ::: No — it only guarantees a critical point. A global maximum can sit at an endpoint? ::: Yes — boundaries are fair game. All-positive Hessian eigenvalues = global min? ::: No — only local min. Flatter minima can generalize better than the deepest one? ::: Yes — lowest training loss isn't the goal.