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It is easy. Observe that $$T[x,t] = \frac{f(x/s)}{s}$$ where $s = \sqrt{t}$ and $f(x) = T[x,1]$. Since $$\int_{\mathbb R} f(x) dx=1 $$ we also have $$\int_{\mathbb R} \frac{f(x/s)}{s} dx=1 $$ simply by means of a trivial change of variables, defining $z = x/s$ where $s>0$. Now take a bounded continuous function $g : \mathbb R \to \mathbb C$, with the aforementioned change of variables we have $$\int_{\mathbb R} \frac{f(x/s)}{s} g(x) dx= \int_{\mathbb R} \frac{f(x/s)}{s} g(sx/s) s dx/s = \int_{\mathbb R} f(z) g(sz) dz\:.$$ Therefore $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) dx = \lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} f(z) g(0) dz = g(0) \int_{\mathbb R} f(z) dz = g(0) 1 = g(0)\:.$$ In other words $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = g(0)\:. \tag{1}$$ The only crucial passage is $$\lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} \lim_{s\to 0^+} f(z) g(sz) dz$$ As quite mild condition that guarantees the passage is that $g$ is bounded as already required (as consequence of Lebesgue's dominated convergence theorem).
I stress that (1) (where $g$ is a smooth compactly supported functions and we obtained the result with much weaker hypotheses) is one of the possible ways to rigorously state that $$T[x,0] = \delta(x)\:.$$ The heat kernel is used to construct the solution of the heat equation $g=g(x,t)$ out of the initial condition $g(x)$: $$g(x,t) = \int_{\mathbb R} T[x-y,t]g(y) dy $$ satisfies $$ \frac{∂g}{∂t} = D \frac{∂^2g}{∂x^2} $$ for $t>0$ with initial condition $$g(x,0) = g(x)\:.$$ The proof is immediate from (1).

It is easy. Observe that $$T[x,t] = \frac{f(x/s)}{s}$$ where $s = \sqrt{t}$ and $f(x) = T[x,1]$. Since $$\int_{\mathbb R} f(x) dx=1 $$ we also have $$\int_{\mathbb R} \frac{f(x/s)}{s} dx=1 $$ simply by means of a trivial change of variables, defining $z = x/s$ where $s>0$. Now take a bounded continuous function $g : \mathbb R \to \mathbb C$, with the aforementioned change of variables we have $$\int_{\mathbb R} \frac{f(x/s)}{s} g(x) dx= \int_{\mathbb R} \frac{f(x/s)}{s} g(sx/s) s dx/s = \int_{\mathbb R} f(z) g(sz) dz\:.$$ Therefore $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) dx = \lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} f(z) g(0) dz = g(0) \int_{\mathbb R} f(z) dz = g(0) 1 = g(0)\:.$$ In other words $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = g(0)\:. \tag{1}$$ The only crucial passage is $$\lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} \lim_{s\to 0^+} f(z) g(sz) dz$$ A quite mild condition that guarantees the passage is that $g$ is bounded as already required (as consequence of Lebesgue's dominated convergence theorem).
I stress that (1) (where $g$ is a smooth compactly supported functions and we obtained the result with much weaker hypotheses) is one of the possible ways to rigorously state that $$T[x,0^+] = \delta(x)\:.$$ The heat kernel is used to construct the solution of the heat equation $g=g(x,t)$ out of the initial condition $g(x)$: $$g(x,t) = \int_{\mathbb R} T[x-y,t]g(y) dy $$ satisfies $$ \frac{∂g}{∂t} = D \frac{∂^2g}{∂x^2} $$ for $t>0$ with initial condition $$g(x,0) = g(x)\:.$$ The proof is immediate from (1).

It is easy. Observe that $$T[x,t] = \frac{f(x/s)}{s}$$ where $s = \sqrt{t}$ and $f(x) = T[x,1]$. Since $$\int_{\mathbb R} f(x) dx=1 $$ we also have $$\int_{\mathbb R} \frac{f(x/s)}{s} dx=1 $$ simply by means of a trivial change of variables, defining $z = x/s$ where $s>0$. Now take a bounded continuous function $g : \mathbb R \to \mathbb C$, with the aforementioned change of variables we have $$\int_{\mathbb R} \frac{f(x/s)}{s} g(x) dx= \int_{\mathbb R} \frac{f(x/s)}{s} g(sx/s) s dx/s = \int_{\mathbb R} f(z) g(sz) dz\:.$$ Therefore $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = \lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} f(z) g(0) dz = g(0) \int_{\mathbb R} f(z) dz = g(0) 1 = g(0)\:.$$ In other words $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = g(0)\:. \tag{1}$$ The only crucial passage is $$\lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} \lim_{s\to 0^+} f(z) g(sz) dz$$ As quite mild condition that guarantees the passage is that $g$ is bounded as already required (as consequence of Lebesgue's dominated convergence theorem).
I stress that (1) (where $g$ is a smooth compactly supported functions and we obtained the result with much weaker hypotheses) is one of the possible ways to rigorously state that $$T[x,0] = \delta(x)\:.$$ The heat kernel is used to construct the solution of the heat equation $g=g(x,t)$ out of the initial condition $g(x)$: $$g(x,t) = \int_{\mathbb R} T[x-y,t]g(y) dy $$ satisfies $$ \frac{∂g}{∂t} = D \frac{∂^2g}{∂x^2} $$ for $t>0$ with initial condition $$g(x,0) = g(x)\:.$$ The proof is immediate from (1).

It is easy. Observe that $$T[x,t] = \frac{f(x/s)}{s}$$ where $s = \sqrt{t}$ and $f(x) = T[x,1]$. Since $$\int_{\mathbb R} f(x) dx=1 $$ we also have $$\int_{\mathbb R} \frac{f(x/s)}{s} dx=1 $$ simply by means of a trivial change of variables, defining $z = x/s$ where $s>0$. Now take a bounded continuous function $g : \mathbb R \to \mathbb C$, with the aforementioned change of variables we have $$\int_{\mathbb R} \frac{f(x/s)}{s} g(x) dx= \int_{\mathbb R} \frac{f(x/s)}{s} g(sx/s) s dx/s = \int_{\mathbb R} f(z) g(sz) dz\:.$$ Therefore $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) dx = \lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} f(z) g(0) dz = g(0) \int_{\mathbb R} f(z) dz = g(0) 1 = g(0)\:.$$ In other words $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = g(0)\:. \tag{1}$$ The only crucial passage is $$\lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} \lim_{s\to 0^+} f(z) g(sz) dz$$ As quite mild condition that guarantees the passage is that $g$ is bounded as already required (as consequence of Lebesgue's dominated convergence theorem).
I stress that (1) (where $g$ is a smooth compactly supported functions and we obtained the result with much weaker hypotheses) is one of the possible ways to rigorously state that $$T[x,0] = \delta(x)\:.$$ The heat kernel is used to construct the solution of the heat equation $g=g(x,t)$ out of the initial condition $g(x)$: $$g(x,t) = \int_{\mathbb R} T[x-y,t]g(y) dy $$ satisfies $$ \frac{∂g}{∂t} = D \frac{∂^2g}{∂x^2} $$ for $t>0$ with initial condition $$g(x,0) = g(x)\:.$$ The proof is immediate from (1).

It is easy. Observe that $$T[x,t] = \frac{f(x/s)}{s}$$ where $s = \sqrt{t}$ and $f(x) = T[x,1]$. Since $$\int_{\mathbb R} f(x) dx=1 $$ we also have $$\int_{\mathbb R} \frac{f(x/s)}{s} dx=1 $$ simply by means of a trivial change of variables, defining $z = x/s$ where $s>0$. Now take a bounded continuous function $g : \mathbb R \to \mathbb C$, with the aforementioned change of variables we have $$\int_{\mathbb R} \frac{f(x/s)}{s} g(x) dx= \int_{\mathbb R} \frac{f(x/s)}{s} g(sx/s) s dx/s = \int_{\mathbb R} f(z) g(sz) dz\:.$$ Therefore $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = \lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} f(z) g(0) dz = g(0) \int_{\mathbb R} f(z) dz = g(0) 1 = g(0)\:.$$ In other words $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = g(0)\:. \tag{1}$$ The only crucial passage is $$\lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} \lim_{s\to 0^+} f(z) g(sz) dz$$ As quite mild condition that guarantees the passage is that $g$ is bounded as already required (as consequence of Lebesgue's dominated convergence theorem).
I stress that (1) (where $g$ is a smooth compactly supported functions and we obtained the result with much weaker hypotheses) is one of the possible ways to rigorously state that $$T[x,0] = \delta(x)\:.$$

It is easy. Observe that $$T[x,t] = \frac{f(x/s)}{s}$$ where $s = \sqrt{t}$ and $f(x) = T[x,1]$. Since $$\int_{\mathbb R} f(x) dx=1 $$ we also have $$\int_{\mathbb R} \frac{f(x/s)}{s} dx=1 $$ simply by means of a trivial change of variables, defining $z = x/s$ where $s>0$. Now take a bounded continuous function $g : \mathbb R \to \mathbb C$, with the aforementioned change of variables we have $$\int_{\mathbb R} \frac{f(x/s)}{s} g(x) dx= \int_{\mathbb R} \frac{f(x/s)}{s} g(sx/s) s dx/s = \int_{\mathbb R} f(z) g(sz) dz\:.$$ Therefore $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = \lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} f(z) g(0) dz = g(0) \int_{\mathbb R} f(z) dz = g(0) 1 = g(0)\:.$$ In other words $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = g(0)\:. \tag{1}$$ The only crucial passage is $$\lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} \lim_{s\to 0^+} f(z) g(sz) dz$$ As quite mild condition that guarantees the passage is that $g$ is bounded as already required (as consequence of Lebesgue's dominated convergence theorem).
I stress that (1) (where $g$ is a smooth compactly supported functions and we obtained the result with much weaker hypotheses) is one of the possible ways to rigorously state that $$T[x,0] = \delta(x)\:.$$ The heat kernel is used to construct the solution of the heat equation $g=g(x,t)$ out of the initial condition $g(x)$: $$g(x,t) = \int_{\mathbb R} T[x-y,t]g(y) dy $$ satisfies $$ \frac{∂g}{∂t} = D \frac{∂^2g}{∂x^2} $$ for $t>0$ with initial condition $$g(x,0) = g(x)\:.$$ The proof is immediate from (1).