You seem to be confused about what you want. If you want to decrease the motor speed, but you still want maximum torque, then you must apply full rated electrical power to the motor, and put a mechanical brake on the motor until it slows to the speed you desire. Or, you must somehow make your motor less efficient. I don't think that's what you want.
Think of it this way: electrical power is the product of current \$I\$ and voltage \$E\$:
$$ P = I E $$
Mechanical power is the product of torque (\$\tau\$, in newton-meters) in and angular velocity (\$\omega\$, in radians per second):
$$ P = \tau \omega $$
A motor is an electrical to mechanical power converter. The mechanical power always equals the electrical power after losses.
Furthermore, current is proportional to torque, because the more current you apply, the stronger the magnetic field inside the motor, and the attraction between the motor's poles becomes greater.
If the mechanical and electrical powers are correlated, as are the current and torque, then voltage and speed must be, also. And they are, because the faster the rotor spins through the stator field, the greater back-emf it will generate. This is Faraday's law of induction.
So, if you want to decrease speed, decrease voltage. If you want to decrease torque, decrease current. If you increase torque (say by putting a brake on the motor), you are increasing motor torque. But if you don't change the supply of electrical power, then the mechanical power also won't change. If torque increased, the only way to keep mechanical power constant is to decrease speed, so the motor slows down.
There is one kink here: as torque goes up, current goes up. The resistive losses in the motor also go up, because the windings have some resistance, and those resistive losses are proportional to the square of the current:
$$ P = R I^2 $$
So, as current goes up, the resistive losses increase, making the motor a less efficient converter of electrical energy to mechanical energy, because some of that electrical energy is now creating heat. If you stall the motor, then the motor reaches 0% efficiency: speed is zero, so mechanical power must be zero, but the motor is drawing a ton of current, and there is a voltage drop over the winding resistance, so electrical power is very high.
Interesting fact: if you can make a motor with no winding resistance (or other losses), and you connect it to a perfect voltage source, then the speed regulation (how much speed changes with torque) is perfect. That is, the motor won't slow down if you try to stop it: it will just draw exactly enough more current from your battery to keep spinning at the same speed, no matter what.
PWM is all irrelevant to this. PWM motor control is just a way to efficiently apply less than the full battery voltage to the motor. It works because a PWM driven motor is equivalent to a buck converter. Changing your PWM duty cycle is equivalent to changing your supply voltage:
The maximum torque you could have (which you will get when the motor is stalled) is limited by the current your power supply can supply and the losses in the motor, just as it is without PWM. Your PWM driver might add a bit of resistance to the circuit, reducing the current and torque a bit, but usually this isn't significant compared to the resistance of the motor windings.
I have a section here in my lecture notes that's implying the startup torque is equal to the loading torque but that makes no sense to me.
And you are right. The airgap torque has to be higher than the load torque in order for the machine to accelerate. Alternatively, the airgap torque has to be lower than the load torque for the machine to accelerate in the backward direction.
Just the fact that the motor is at standstill does not imply zero torque. It implies that the load torque is equal to the airgap torque, zero or non-zero.
The static load could be X [Nm] and the airgap torque could cancel it out if it were also X [Nm]. Many loads have the load torque proportional to the square of the shaft speed. Hence, the load torque is zero at standstill. This means that your DC motor can develop any torque equal to \$K_t I_{winding}\$ up to rated winding current \$I_{rated}\$. The startup torque of the motor would be equal to \$T_{start} = V_{dc}/R_{winding} * K_t\$ [Nm]. Note that the current \$I_{winding}\$ would likely be higher than the rated current if not actively controlled.
If the motor windings are initially disconnected or shorted (provided non-zero winding resistance) and the shaft speed is zero, then such state implies zero load torque.
Best Answer
Just like it's simplest to learn about a lossless inductor first, so let's start with more or less lossless motor. We'll take account of losses when we have to, but they are not essential for basic understanding.
A motor is also a generator. Spin it, and it generates volts on the armature. It doesn't matter whether it's spinning because it's a motor, or spinning because you're driving it as a generator, speed = armature_volts/k.
Pass a current through it and it generates a torque. Torque = armature_current.k
You can think of a motor as a mechanical transformer. Power in = power out. Volts x amps in = speed x torque out. When equating power, that pesky k has cancelled out. If you change the value of k, the torque constant, then the motor gets faster and delivers less torque, or vice versa, but the power balance is the same. If you run the same machine as a generator, then speed x torque in = volts x amps out.
What happens if you apply a voltage source to a motor at rest?
Two things happen, at different speeds, the first so quickly you may not notice, the second rather more slowly.
A motor armature has inductance, Larm, and resistance Rarm. At the moment of switch on, we apply V to the armature. The current starts to increase. It initially increases at such a rate that Larm x dI/dt generates a back EMF equal to the terminal voltage. The current flowing through Rarm generates a voltage IRarm which opposes V, so there is less voltage across the inductance to drive an increase of current, so the rate of current increase slows down. Eventually, the current has increased to settle at V/Rarm, with a time constant of Larm/Rarm, typically in a matter of mS.
With a 'good' motor with a low Rarm, this current will typically be very large. It's known as the 'starting' current, for obvious reasons. Small motors are rated to be started like this safely. Big motors cannot be started like this, and need some sort of soft start controller.
So far, the motor still hasn't moved, the mechanical inertia means it's not rotating yet, or has barely started. But now there is a big armature current flowing, which generates a torque, and the motor accelerates.
Once it is turning, at any speed, it generates a back EMF proportional to its speed. This back EMF reduces the effective terminal voltage available to drive current through Rarm. The armature current therefore falls (with a time constant of Larm/Rarm), and so generates less torque.
Eventually the motor reaches a balance, where it's at a speed where the generated back EMF balances off most of the input voltage, and the small difference in voltage that remains drives an armature current through Rarm, which generates enough torque in the motor to match the load torque, plus loss torques like friction and air resistance.