Show that the sum of a function series is bounded

Question

For every $k\in\mathbb{N}$, with $k\geq 1$, let $f_k=f_k(x,t)$ be the real-valued function defined over the set $(x,t) \in [-\pi,\pi]\times[0,+\infty)$ by

$$f_k(x,t)=(-1)^{k+1}\frac{2}{k}e^{-k^{2} t}\sin(kx).$$

We then consider the corresponding function series

$$\sum_{k=1}^{+\infty}f_k(x,t)=\sum_{k=1}^{+\infty}(-1)^{k+1}\frac{2}{k}e^{-k^{2} t}\sin(kx). \tag{1}$$

It is simple to show that $(1)$ converges pointwise on the whole $[-\pi,\pi]\times[0,+\infty)$, and uniformly on every $[-\pi,\pi]\times[t_0,+\infty)$, with $t_0>0$. Let $u=u(x,t)$ be its sum, i.e

$$u(x,t)=\sum_{k=1}^{+\infty}f_k(x,t)=\sum_{k=1}^{+\infty}(-1)^{k+1}\frac{2}{k}e^{-k^{2} t}\sin(kx).$$

The sum $u$ it is a superposition of sinusoids of increasing frequency $\frac{k}{2\pi}$ and of strongly damped amplitude because of the negative exponential, at least when $t > 0$. For this reason, it is simple to show that $u$ is smooth on the set $[-\pi,\pi]\times(0,+\infty)$, ie $u\in C^{\infty}([-\pi,\pi]\times(0,+\infty))$. Notice also that:

1. $u=u(x,t)$ is the solution of the one-dimensional Heat Equation problem with periodic boundary conditions

$$\begin{cases} u_t-u_{xx} = 0 \qquad &x \in (-\pi,\pi),t>0 \\ u(x,0) = x \qquad &x \in (-\pi,\pi) \\ u(-\pi,t) = u(\pi,t) \qquad &t \geq 0 \end{cases}. \tag{2}$$

2. For every $x_0\in (-\pi,\pi)$ one has $$\lim_{(x,t)\to(x_0,0)}u(x,t)=x_0,$$ and then $u$ is also continuous at every point of the open segment $(-\pi,\pi)\times\{0\}$.

3. The limits $$\lim_{(x,t)\to(\pm \pi,0)}u(x,t)\qquad \nexists.$$

I'm not able to prove that $u$ is bounded on the whole $[-\pi,\pi]\times [0,+\infty)$.

By uniform convergence (as suggested me in comments), we just need to prove that partial sums of $(1)$ are uniformly bounded on $[-\pi,\pi]\times [0,+\infty)$, but I really don't know how to obtain this uniform bound.

Any hint would be really appreciated.

Answer 1

We use that $\left|\sum_{k=1}^{n}\frac{(-1)^ksin(kx)}{k}\right|\leq C$ which is a classic result; see for example this MSE post

(edited per comments $\sum_{k=1}^{n}\frac{(-1)^{k+1} sin(kx)}{k}= \sum_{k=1}^{n}\frac{sin(kx)}{k}- \sum_{k=1}^{[n/2]}\frac{sin(k(2x))}{k}$, so the bound for the signed sum follows from the bound from the unsigned sum applied for $x, 2x$)

and summation by parts, namely if $\sum_{k=1}^n a_k =A_n$ then

$\sum_{k=1}^n a_kb_k=A_1(b_1-b_2)+A_2(b_2-b_3)+...A_{n-1}(b_{n-1}-b_n)+A_nb_n$ so if

$|A_n| \le C, b_1 \ge b_2 \ge..b_n \ge 0$ we get:

$|\sum_{k=1}^n a_kb_k| \le C(b_1-b_2)+...Cb_n=Cb_1$

In our case since $a_k =(-1)^{k+1}\frac{2}{k}sin(kx), b_k =e^{-k^2t}$ obviously satisfy the hypothesis above for $x \in R, t \ge 0$ we get:

$\left|\sum_{k=1}^{n}(-1)^{k+1}\frac{2}{k}e^{-k^2t}sin(kx)\right|\leq 2Ce^{-t^2} \le 2C$ uniformly in $(x,t) \in [-\pi,\pi] \times [0, \infty)$

Answer 2

(ANSWER)

Theorem. (Konard Knopp. "Theory and Applications of Infinite Series".Dover.1990.pg.348) The series $\sum a_{\nu}(t)b_{\nu}(t)$ is uniformly convergent in $J$ if the series $\sum |b_{\nu}-b_{\nu+1}|$ converges uniformly in $J$, and the series $\sum a_{\nu}$ has uniformly bounded partial sums, provided the functions $b_{\nu}(t)\rightarrow 0$ uniformly in $J$.

Let $\delta$ be any fixed number of $(0,1/2)$. Fix also $x\in(-\pi,\pi)$. Set $$ b_{\nu}=\frac{1}{\nu^{1/2+\delta}}e^{-\nu^2t} $$ and $$ a_{\nu}=\frac{2(-1)^{\nu+1}}{\nu^{1/2-\delta}}\sin(\nu x). $$ Then clearly for $t\geq 0$ $$ \sum^{\infty}_{\nu=1}\left|\frac{e^{-\nu^2 t}}{\nu^{1/2+\delta}}-\frac{e^{-(\nu+1)^2t}}{(\nu+1)^{1/2+\delta}}\right|=e^{-t}<\infty(uniformly). $$ and $b_{\nu}(t)\rightarrow 0$. Also if $x=y-\pi$, then $$ \sum_{1\leq\nu\leq M}\frac{2(-1)^{\nu+1}}{\nu^{1/2-\delta}}\sin(\nu x)=-2\sum_{1\leq\nu\leq M}\frac{1}{\nu^{1/2-\delta}}\sin(\nu y)\tag 1 $$ is uniformly bouded in $\textbf{R}$. This last argument follows from the well known fact that if $c_{\nu}$ is null and monotone then $\sum c_{\nu}\sin(\nu y)$ is uniformly convergent everywhere in $\epsilon\leq y\leq 2\pi-\epsilon$, $0<\epsilon<\pi$ Or equivalently $\epsilon-\pi\leq x\leq \pi-\epsilon$ (see the above reference pg.349). The cace $x=\pm\pi$ is trivial. QED

Answer 3

Firstly,

$f_k = -\dfrac2ke^{-k^2t}\sin k(x+\pi) = -\dfrac2ke^{-k^2t}\sin ky(x),$

where

$y(x)= \text{ mod }(x+\pi, 2\pi) - \pi.$

At the same time:

• Since $\;\forall(k\in\mathbb N)\forall(x\in[-\pi,\pi]):\; \left|-\dfrac2k\sin k(x+\pi)\right|\le 2,\;$ then the function $\;g_k(x) = -\dfrac2k\sin k(x+\pi)\;$ is bounded in $\;[-\pi,\pi];\;$
• $\forall(k\in\mathbb N)\forall(x\in\{-\pi,\pi\} \;\exists\left(\lim\limits_{\varepsilon\to -0}\,\sum\limits_{k=1}^\infty g_k(x+\varepsilon) = \lim\limits_{\varepsilon\to +0} \sum\limits_{k=1}^\infty g_k(x+\varepsilon)=(\pi-|y(x)|)\text{ sgn }y(x) = \Phi(x)\right)\;$
  (see also WA result and checking);
• Since $\;\forall(k\in\mathbb N)\forall(t\in[0,\infty))\; e^{-k^2t}\in [0,1],\;$ then the function $\;h_k(t)=e^{-k^2t}\;$ is bounded in $\;[0,\infty);$
• $\forall(k\in\mathbb N)\forall(t\in(0,\infty)\;\exists\left(\lim\limits_{\varepsilon\to -0}\,\sum\limits_{k=1}^\infty h_k(t+\varepsilon) = \lim\limits_{\varepsilon\to +0} \sum\limits_{k=1}^\infty h_k(t+\varepsilon)=\Psi(t),\quad |\Psi(t)|\le \sum\limits_{k=1}^\infty e^{-kt}\le \dfrac {e^{-t}}{1-e^{-t}} =\dfrac1{e^t-1}\right);$
• Functions $\;g_k(x), \Phi(x), h_k(t), \Psi(t)$ are continuous in their domains.

Then

• If $\;t\in(0,\infty)\;$ then $|u(x,t)| \le \sum\limits_{k=1}^\infty |g_k(x)|\,h_k(t) \le 2\sum\limits_{k=1}^\infty h_k(t) \le\dfrac2{e^t-1};$
• If $\;t=0,\;$ then $|u(x,0)| \le |\Phi(x)| \le \pi.$

On the other hand, $$\sum\limits_{k=1}^n g^2_k(x)\le \sum\limits_{k=1}^n\dfrac1{k^2}\le H^{(2)}_n,$$ $$\sum\limits_{k=1}^n h^2_k(t)\le \sum\limits_{k=1}^n e^{-2k^2}\le\dfrac{1-e^{-(2n+2)t}}{1-e^{-2t}}.$$

Taking in account Cauchi-Schwartz inequality, easily to get $$\left|\sum\limits_{k=1}^n g_k h_k\right|\le \sqrt{H^{(2)}_n\,\dfrac{1-e^{-(2n+2)t}}{1-e^{-2t}}}.$$

Therefore, $\;u(x,t)\;$ is uniformly bounded in $\;\color{brown}{\mathbf{\left([-\pi,\pi]\times(0,\infty)\right).}}$