Let $\theta \in \ell^2(\mathbb{N})$ satisfy $$ \sum_{j=1}^n \theta_j^2 a_j^2 \leq Q $$ for some $Q > 0$ where $$ a_j = \begin{cases} j^\beta & \text{$j$ even}\\ (j-1)^\beta & \text{$j$ odd} \end{cases} $$ for some $\beta > 1/2$. Let $\{\varphi_j\}_{j=1}^\infty$ denote the trigonometric basis of $[0,1]$, i.e. $\varphi_1(x) = 1$, $\varphi_{2k}(x) = \sqrt{2} \cos(2\pi k x)$ and $\varphi_{2k+1}(x) = \sqrt{2} \sin(2\pi k x)$ for $k \in \mathbb{N}$. Show that $f = \sum_{j=1}^\infty \theta_j \varphi_j$ is continuous.
I know that $\varphi_j$ is Lipschitz with constant $\pi\lfloor j+1 \rfloor$ but I don't think that helps very much. Any ideas?
Note first that the hypothesis is just a silly way of writing $$\sum_{j=1}^\infty\theta_j^2j^{2\beta}<\infty.$$
Proof by Looking it Up
The hypothesis says precisely that $f$ is in the Sobolev space often denoted $H^\beta$; since $\beta>1/2$ the Sobolev Embedding Theorem shows that $f$ is continuous.
Actual Proof
$$\sum|\theta_j|=\sum|\theta_j|j^\beta j^{-\beta}\le\left(\sum\theta_j^2j^{2\beta}\right)^{1/2}\left(\sum j^{-2\beta}\right)^{1/2}<\infty,$$since $2\beta>1$. So the series converges uniformly, hence $f$ is continuous.
Proof of a Stronger Result
Say $x<y$ and $y-x$ is small. Choose an integer $N$ with $1/2\le N(y-x)\le 2$. Now $$f(y)-f(x)=\left(\sum_{j=1}^N+\sum_{j=N+1}^\infty\right)\theta_j(\phi_j(y)-\phi_j(x)):=I+II.$$Now the Lipschitz constant for $\phi_j$ shows that $$I\le c(y-x)\sum_{j=1}^N\theta_jj^\beta j^{1-\beta}\le c(y-x)\left(\sum_{j=1}^N\theta_j^2j^{2\beta}\right)^{1/2}\left(\sum_{j=1}^Nj^{2-2\beta}\right)^{1/2}\le c(y-x)N^{\frac32-\beta}\le c(y-x)^{\beta-\frac12}.$$To estimate $II$ you just use the fact that $|\phi_j|\le1$: $$II\le 2\sum_{j=N+1}^\infty\theta_j j^\beta j^{-\beta}\le c\left(\sum_{j=N+1}^\infty j^{-2\beta}\right)^{1/2}\le cN^{\frac12-\beta}\le c(y-x)^{\beta-\frac12}.$$Since $\beta-\frac12>0$ this shows that $f\in Lip_{\beta-\frac12}$.
Note
It's worthwhile spending some time understanding this argument, because the outline "use a trivial uniform estimate for the high-order terms and the Lipschitz condition on the low-order terms" comes up a lot in this sort of thing.