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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 $n \leq C^2$. |
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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 $n \leq C^2$. |
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