Functional Derivative ${\delta q_a(t)}/{\delta q_b(t')}$ $\newcommand{\fdv}[2]{\frac{\delta #1}{\delta #2}}$
$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}$
$\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$
I'm from a physics background and I've always known the definition to be related to the Euler-Lagrange Equations i.e.
$$\fdv {L(q,\dot q)} {q(t)} = \pdv{L}{q} - \dv{}{t} \pdv{L}{\dot q} \; ,$$
where $\dot q$ denotes the derivative of the function $q$ with respect to $t$.
However with this definition I cannot prove a fact that I've seen in a lecture video about Quantum Field Theory, which is
$$\fdv{q_a(t)}{q_b(t')} = \delta_{ab} \delta(t-t') \; , \tag 1$$
where $\delta(t-t')$ is the Dirac delta function/distribution and $\delta_{ab}$ is the Kroneker delta.
I don't know how I should make sense out of this equation let alone prove it because $q_a(t)$ is not a functional but a function. I'd appreciate if someone can explain what is meant by the equation (1) and give a proof thereof.
 A: First have a look at:
Integral Argument

For enough variations:
  $$\varphi\in\mathcal{C}^\infty:\quad\int_0^T\varphi(s)\delta\varphi(s)\mathrm{d}s=0\quad(\delta\varphi\in\mathcal{C}^\infty_0)\implies\varphi(t)\equiv0$$

That is needed for Euler-Lagrange!
Now go carefully through:
Detailed Calculation

Directional derivative:
  $$\frac{\delta}{\delta q}:=\frac{\partial}{\partial\delta q}:=\partial_{\delta q}$$

Consider as example:
$$\mathcal{L}:\mathbb{R}^3\to\mathbb{R}:\quad\mathcal{L}(u,v;w):=\frac12mv^2-\frac{1}{w}u^2$$
For partial derivatives:
$$\frac{\partial\mathcal{L}}{\partial u}=-\frac{1}{w}2u\quad\frac{\partial\mathcal{L}}{\partial v}=mv\quad\frac{\partial\mathcal{L}}{\partial w}=\frac{1}{w^2}u^2$$
Introduce evaluations:
$$\eta_a:\mathcal{C}^\infty(\mathbb{R})\to\mathbb{R}:\quad\eta_a[q]:=q(a)$$
$$\vartheta_a:\mathcal{C}^\infty(\mathbb{R})\to\mathbb{R}:\quad\vartheta_a[q]:=\dot{q}(a)$$
$$\omega_a:\mathcal{C}^\infty(\mathbb{R})\to\mathbb{R}:\quad\omega_a[q]:=a$$
For directional derivative:
$$\frac{\delta\eta_a}{\delta q}[q]=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{\eta_a[q+\varepsilon\delta q]-\eta_a[q]\}=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{q(a)+\varepsilon\delta q(a)-q(a)\}=\delta q(a)$$
$$\frac{\delta\vartheta_a}{\delta q}[q]=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{\vartheta_a[q+\varepsilon\delta q]-\vartheta_a[q]\}\stackrel{*}{=}\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{\dot{q}(a)+\varepsilon\dot{\delta q}(a)-\dot{q}(a)\}=\dot{\delta q}(a)$$
$$\frac{\delta\omega_a}{\delta q}[q]=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{\omega_a[q+\varepsilon\delta q]-\omega_a[q]\}=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{a-a\}=0$$
Consider the action:
$$S[q]:=\int_0^T\mathcal{L}(\eta_s[q],\vartheta_s[q];s)\mathrm{d}s=:\int_0^T\hat{\mathcal{L}_s}[q]\mathrm{d}s\quad(q\in\mathcal{C}^\infty)$$
Suppose variations:
$$\mathcal{C}^\infty_0:=\{\delta q\in\mathcal{C}^\infty:\delta q(0)=\delta q(T)=0\}$$
For directional derivative:
$$\frac{\delta S}{\delta q}[q]=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{S[q+\varepsilon\delta q]-S[q]\}\\
=\lim_{\varepsilon\to0}\int_0^T\frac{1}{\varepsilon}\bigg\{\mathcal{L}(\eta_s[q+\varepsilon\delta q],\vartheta_s[q+\varepsilon\delta q];s)\mathrm{d}s-\mathcal{L}(\eta_s[q],\vartheta_s[q];s)\bigg\}\mathrm{d}s\\
\stackrel{**}{=}\int_0^T\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\bigg\{\mathcal{L}(\eta_s[q+\varepsilon\delta q],\vartheta_s[q+\varepsilon\delta q];s)\mathrm{d}s-\mathcal{L}(\eta_s[q],\vartheta_s[q];s)\bigg\}\mathrm{d}s\\
=\int_0^T\frac{\delta\hat{\mathcal{L}_s}}{\delta q}[q]\mathrm{d}s=\int_0^T\bigg\{\frac{\partial\mathcal{L}}{\partial u}\frac{\delta\eta_s}{\delta q}+\frac{\partial\mathcal{L}}{\partial v}\frac{\delta\vartheta_s}{\delta q}+\frac{\partial\mathcal{L}}{\partial w}\frac{\delta\omega_s}{\delta q}\bigg\}\mathrm{d}s\\
=\int_0^T\bigg\{\frac{\partial\mathcal{L}}{\partial u}\delta q(s)+\frac{\partial\mathcal{L}}{\partial v}\dot{\delta q}(s)+0\bigg\}\mathrm{d}s=\int_0^T\bigg\{\frac{\partial\mathcal{L}}{\partial u}\delta q(s)+\frac{\partial\mathcal{L}}{\partial v}\frac{\mathrm{d}}{\mathrm{d}s}\delta q(s)\bigg\}\mathrm{d}s\\
=\int_0^T\bigg\{\frac{\partial\mathcal{L}}{\partial u}\delta q(s)-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial v}\delta q(s)\bigg\}\mathrm{d}s=\int_0^T\bigg\{\frac{\partial\mathcal{L}}{\partial u}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial v}\bigg\}\delta q(s)\mathrm{d}s$$
By the integral argument:
$$\frac{\delta S}{\delta q}[q]=0\quad(\delta q\in\mathcal{C}^\infty_0)\implies\bigg\{\frac{\partial\mathcal{L}}{\partial u}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial v}\bigg\}=0$$
Concluding Euler-Lagrange.
Next try it sombolically:
Symbolic Calculation
Regard action functional:
$$S[q]:=\int\mathcal{L}(q(s),\dot{q}(s);s)\mathrm{d}s\quad(q\in\mathcal{C}^\infty)$$
Its derivite writes:
$$\frac{\delta S}{\delta q}=\int\bigg\{\frac{\partial\mathcal{L}}{\partial q}\frac{\delta q}{\delta q}+\frac{\partial\mathcal{L}}{\partial\dot{q}}\frac{\delta\dot{q}}{\delta q}+\frac{\partial\mathcal{L}}{\partial t}\frac{\delta t}{\delta q}\bigg\}\delta q\mathrm{d}s\\
=\int\bigg\{\frac{\partial\mathcal{L}}{\partial q}\delta q+\frac{\partial\mathcal{L}}{\partial\dot{q}}\dot{\delta q}+\frac{\partial\mathcal{L}}{\partial t}0\bigg\}\delta q\mathrm{d}s=\int\bigg\{\frac{\partial\mathcal{L}}{\partial q}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial\dot{q}}\bigg\}\delta q\mathrm{d}s$$
By the integral argument:
$$\frac{\delta S}{\delta q}=0\quad(\delta q\in\mathcal{C}^\infty_0)\implies\frac{\delta\mathcal{L}}{\delta q}=\frac{\mathrm{d}}{\mathrm{d}s}\frac{\delta\mathcal{L}}{\delta q}$$
Concluding Euler-Lagrange.
Finally try to shorten it by:
Shortened Calculation
Regard action functional:
$$S[q]:=\int\mathcal{L}(q(s),\dot{q}(s);s)\mathrm{d}s\quad(q\in\mathcal{C}^\infty)$$

For singular variations:
  $$\delta q_a\notin\mathcal{C}^\infty_0:\quad\delta q_a(t):=\delta(t-a)$$

Agree on the convention:
$$\frac{\delta S}{\delta q(a)}:=\frac{\delta S}{\delta q_a}\quad\frac{\partial\mathcal{L}}{\partial q(t)}:=\frac{\partial\mathcal{L}}{\partial q}(t)\quad\frac{\partial\mathcal{L}}{\partial\dot{q}(t)}:=\frac{\partial\mathcal{L}}{\partial\dot{q}}(t)\quad\frac{\delta q(t)}{\delta q(a)}:=\frac{\delta q}{\delta q_a}(t)$$
By the previous calculation:
$$0=\frac{\delta S}{\delta q(t)}=\int\bigg\{\frac{\partial\mathcal{L}}{\partial q(s)}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial\dot{q}(s)}\bigg\}\frac{\delta q(s)}{\delta q(t)}\mathrm{d}s\\
=\int\bigg\{\frac{\partial\mathcal{L}}{\partial q(s)}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial\dot{q}(s)}\bigg\}\delta(s-t)\mathrm{d}s=\frac{\partial\mathcal{L}}{\partial q(t)}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial\mathcal{L}}{\partial\dot{q}(t)}$$
That killed the integral argument!
...now you're ready for QFT. ;)
*Linearity of derivative!
**Regularity of Lagrangian!
A: Fix $t=t_0$, so you can pointwise identify $q_a(t_0)$ with a functional $$\mathcal F[u]=\int_{\mathbb R}\mathrm d t' u(t')\delta(t_0-t')$$ as $q_a(t_0)\equiv\mathcal F[q_a]$. Then you have the "Euler-Lagrange equation": $$\frac{\delta \mathcal F[q_a]}{\delta q_b(t')}=\frac{\partial}{\partial q_b(t')}q_a(t')\delta(t_0-t')$$
the same as for a Lagrangian (the $d/dt\dots$ part is of course missing). Now finally, using $\partial q_a(t')/\partial q_b(t')=\delta_{ab}$, the result follows.
