Suppose I have the problem $$u_t=ku_{xx}$$ $x \in \mathbb{R}$, $t>0$ $$u(x,0)=f(x)$$ $x \in \mathbb{R}$ for a square integrable function $f(x)$. Prove $$\max_{x \in \mathbb{R} } | u(x,t) | \leq c t^{-1/4} \| f(x) \|$$ for some positive constant $c$.
My attempt so far, using Cauchy Schwartz inequality I get $$\max_x | u |= \max | \frac{1}{(4k \pi t)^{1/2}} \int_{ - \infty }^{\infty } e^{-\frac{(x-y)^2}{4k t}} f(y) dy | \leq \max_x \frac{1}{(4k \pi t)^{1/2}} \left( \int_{ - \infty }^{\infty } e^{-\frac{(x-y)^2}{2k t}} \right)^{1/2} \| f(x) \| $$ I know that $$\max_x \frac{1}{(4k \pi t)^{1/2}} e^{-\frac{x^2}{4k t}} =\frac{1}{(4k \pi t)^{1/2}} $$ and $$|u_x| \leq A t^{-3/4} \|f(x)|$$ for some other positive constant $A$.
I tried to use those relations but couldn't come up with anything, perhaps if I could explicitly compute $\max_x \frac{1}{(4k \pi t)^{1/2}} \left( \int_{ - \infty }^{\infty } e^{-\frac{(x-y)^2}{2k t}} \right)^{1/2}$ I may end up with something, but I couldn't find anything useful on how to make this computation, so how can I prove this?