Here is another Prelim problem from Advanced Calculus.
For $t>0$ and $D>0$ define $g(x,t)$ by $$ g(x,t)=\frac{1}{\sqrt{Dt}}\exp{\frac{-x^2}{4Dt}} $$ Now, for $f:\mathbf{R}\to\mathbf{R}$ being continuous with compact support, define $u(x,t)$ as $$ u(x,t) = \int_{-\infty}^\infty g(x-y,t)f(y)dy $$ Now show for a fixed $x$ (also find C) for $t\to\infty$ $$ u(x,t) = \frac{C}{\sqrt{t}}+O(\frac{1}{t}). $$ This is the third part of the problem, so maybe it would help to know that $$ \frac{\partial u}{\partial t} = D \frac{\partial^2 u}{\partial x^2}. $$
So my plan of attack was to show, with $C = \frac{1}{\sqrt{D}}\int_{-\infty}^\infty f(y)dy$ $$ |\sqrt{t} u(x,t) - C| \leq M \frac{1}{\sqrt{t}} $$ or try to show $$ \frac{1}{\sqrt{D}}\int_{-\infty}^\infty|\exp{\frac{-(x-y)^2}{4Dt}}-1||f(y)|dy $$ but I could not get it to work out.
Hint: You're looking to end up with a sum, even though the integrand is a product. Sometimes you can play tricks with the functions to change a product to a sum (like taking the logarithm etc.), but here the most natural thing to do might be to express one of the functions as its Taylor series, yielding the sought after sum, and coincidentally it fits in well with the O-notation. You know nothing about $f$ so that can't be expanded, while the exponent has a lovely expansion with generous convergence...
It looks like you might get an even better error term though, $O(t^{-3/2})$. It also looks like you intended to write the differential equation differently: $\frac{\partial u}{\partial t} = D \frac{\partial^2 u}{\partial x^2}.$
EDIT
$$ e^x = 1 + x + \frac {x^2} 2 + \cdots = 1 + O(x) \hspace{10pt} \text{as }x\to 0 $$
$$ \begin{align*} u(x,t) &= \int_{-\infty}^\infty \frac 1 {\sqrt{Dt}} e^{-\frac {(x-y)^2} {4Dt}} f(y) dy \\ &= \frac 1 {\sqrt{Dt}} \int_{-\infty}^\infty \left( 1 + O\left(\frac 1 t\right)\right) f(y) dy \hspace{10pt} \text{as } t\to\infty \\ &= \frac 1 {\sqrt{Dt}} \left( 1 + O\left(\frac 1 t\right)\right) \int_{-\infty}^\infty f(y) dy \\ &= \frac C {\sqrt t} + O\left(\frac 1 {t\sqrt t} \right) \end{align*} $$
I used the compactness of the support of $f$ in the fact that the O-notation isn't dependent on $y$.