Let $\phi_t(x)=\Phi(x,t)=(4\pi t)^{-1/2}\exp(\frac{-x^2}{4t})$ be the heat kernel and let $f$ be a bounded Lipschitz function.
Don't ask about the two different $\phi,\Phi$ notations for the heat kernel, there was something wrong in the homework question, and I didn't pay attention to consistency...
To show: The function $$F(x,t):= \begin{cases} \phi_t * f(x),\quad&t>0 \\ f(x) &t=0 \end{cases} $$ is continuous on $\mathbb{R}\times[0,\infty)$ and solves the heat equation on $\mathbb{R}\times(0,\infty)$, that is, $\partial_t F=\partial_x^2 F$, where $$\phi_t * f(x)=\int_{\mathbb R}\phi_t(y)f(x-y)\mathrm dy$$ is the convolution of $\phi_t$ and $f$.
This is a homework question, and I am always writing my homework solutions in $\LaTeX$, so I pretty much copy pasted most of this part...
My attempt: For continuity, it suffices to show that $F$ is well defined for $t>0$ and continuous at $t=0$, since the convolution of continuous functions is again continuous. We know that $(2\pi t)^{-1}\int_{\mathbb{R}}\exp\left(\frac{-y^2}{2}\right)\mathrm{d}y=1$. Therefore, for $t>0$, we have $$F(x,t)=\int_{\mathbb{R}}\phi_t(y)f(x-y) \leq\|f\|_\infty\int_{\mathbb{R}}\phi_t(y)\mathrm{d}y =\|f\|_{\infty}\int_{\mathbb{R}}(4\pi t)^{-1/2}\exp\left(\frac{-x^2}{4t}\right)\mathrm{d}x <\infty$$ Hence $F$ is well defined and continuous for $t>0$. First, notice that by substituting $x=\sqrt{2t}\cdot y$ we get for all $t>0$ $$\int_{\mathbb{R}}\Phi(x,t)\mathrm{d}x =(2\pi t)^{-1}\int_{\mathbb{R}}\exp\left(\frac{-x^2}{2}\right)\mathrm{d}x=1$$ In particular, for all $\varepsilon>0$ we have \begin{align*} |F(x,t)-f(x)| &=\left|\int_{\mathbb{R}}(f(x-y)-f(x))\Phi(y,t)\mathrm{d}y\right| \\ &\leq\int_{\mathbb{R}}|f(x-y)-f(x)|\Phi(y,t)\mathrm{d}y \\ &=\int_{|y|<\varepsilon}|f(x-y)-f(x)|\Phi(y,t)\mathrm{d}y +\int_{|y|\geq\varepsilon}|f(x-y)-f(x)|\Phi(y,t)\mathrm{d}y \\ &\leq\int_{|y|<\varepsilon}|f(x-y)-f(x)|\Phi(y,t)\mathrm{d}y +2\|f\|_\infty\int_{|y|\geq\varepsilon}\Phi(y,t)\mathrm{d}y \\ &\leq\int_{|y|<\varepsilon}L|x-y-x|\Phi(y,t)\mathrm{d}y +2\|f\|_\infty\int_{|y|\geq\varepsilon}\Phi(y,t)\mathrm{d}y \\ &\leq 2L\varepsilon\int_{|y|<\varepsilon}\Phi(y,t)\mathrm{d}y +2\|f\|_\infty\int_{|y|\geq\varepsilon}\Phi(y,t)\mathrm{d}y \end{align*} where $L>0$ is the Lipschitz constant for $f$, i.e. $|f(x)-f(y)|\leq L|x-y|$ for all $x,y\in\mathbb{R}$.
Clearly $2\|f\|_\infty\int_{|y|\geq\varepsilon}\Phi(y,t)\mathrm{d}y\xrightarrow{t\rightarrow 0}0$ for every $\varepsilon>0$. Note that the integral is bounded for every $t>0$. Therefore $$\lim_{t\rightarrow 0}|F(x,t)-f(x)|=0$$ Then $F$ is continuous on $\mathbb{R}\times\{0\}$, hence on all of $\mathbb{R}\times[0,\infty)$.
For $t>0$ we have \begin{align*} \partial_t F(x,t) &=\int_{\mathbb{R}}f(y)\partial_t\Phi(x-y,t)\mathrm{d}y \\ \end{align*} by dominated convergence.
We had a theorem that says if $\phi_t=\Phi(\cdot,t)\in C_b^n(\mathbb R)$ for fixed $t>0$, then $$\partial_x^n(f*\phi_t)=f*(\partial_x^n\phi_t)$$ which is obviously true here. Then, again for all $t>0$ \begin{align*} \partial_x^2 F(x,t) &=\int_{\mathbb{R}}f(x-y)\partial_x^2\Phi(x-y,t)\mathrm{d}y \end{align*} Since $\Phi$ solves the heat equation, $F$ is also a solution to the heat equation.
Is this proof complete and is every step sufficiently justified?
I was having some problems, but as I was writing this post, I got some insights to complete the proof.
Any feedback would be very much appreciated!