Another solution: as Mikael wrote, $||f||_{\infty} \leq C ||f||_2$ for every $f \in E$. Let $f_1,\ldots,f_n$ be an orthonormal family in your subspace. Then for every $x \in [0,1]$, $f_1(x)^2+\ldots+f_n(x)^2 \leq ||f_1(x)f_1+\ldots+f_n(x)f_n||_{\infty} \leq C ||f_1(x)f_1+\ldots+f_n(x)f_n||_2=C \sqrt{f_1(x)^2+\ldots+f_n(x)^2}$, |f_1(x)f_1+\ldots+f_n(x)f_n\|_2$$$=C \sqrt{f_1(x)^2+\ldots+f_n(x)^2},$$ and by squaring we get$f_1(x)^2+\ldots+f_n(x)^2 \leq C^2$, and integrating gives$n \leq C^2$. 1 Another solution: as Mikael wrote,$||f||_{\infty} \leq C ||f||_2$for every$f \in E$. Let$f_1,\ldots,f_n$be an orthonormal family in your subspace. Then for every$x \in [0,1]$,$f_1(x)^2+\ldots+f_n(x)^2 \leq ||f_1(x)f_1+\ldots+f_n(x)f_n||_{\infty} \leq C ||f_1(x)f_1+\ldots+f_n(x)f_n||_2=C \sqrt{f_1(x)^2+\ldots+f_n(x)^2}$, and by squaring we get$f_1(x)^2+\ldots+f_n(x)^2 \leq C^2$, and integrating gives$n \leq C^2\$.