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Let $f:=\varphi$, $a:=\|f\|_2$, $b:=\|f'\|_2$, so that $\|f\|_{H^1}=a+b$; see e.g. Wikipedia for the definition of $H^k$. Without loss of generality, $x=0$. For all $y\in[0,1]$, we have $$|f(y)-f(0)|\le\int_0^y|f'(t)|dt\le\int_0^1|f'(t)|dt\le\sqrt{\int_0^1|f'(t)|^2 dt}\le b,$$ by H"older's inequality, whence $|f(0)|\le b+|f(y)|$. So, \begin{equation} f(0)^2\le\int_0^1(b+|f(y)|)^2dy\le2\int_0^1(b^2+|f(y)|^2)dy\le2(b^2+a^2)\le2(a+b)^2=2\|f\|_{H^1}^2, \end{equation} whence \begin{equation} |f(0)|\le\sqrt2\,\|f\|_{H^1}, \end{equation} as desired.

Using here the interval $[0,a/b]$ instead of $[0,1]$, one can improve the above inequality to \begin{equation} |f(0)|\le2\sqrt{ab}=2\sqrt{\|f\|_2\,\|f'\|_2}\le \|f\|_{H^1}. \end{equation}

Let $f:=\varphi$, $a:=\|f\|_2$, $b:=\|f'\|_2$, so that $\|f\|_{H^1}=a+b$; see e.g. Wikipedia for the definition of $H^k$. Without loss of generality, $x=0$. For all $y\in[0,1]$, we have $$|f(y)-f(0)|\le\int_0^y|f'(t)|dt\le\int_0^1|f'(t)|dt\le\sqrt{\int_0^1|f'(t)|^2 dt}\le b,$$ by H"older's inequality, whence $|f(0)|\le b+|f(y)|$. So, \begin{equation} \begin{split} f(0)^2\le\int_0^1(b+|f(y)|)^2dy& \le2\int_0^1(b^2+|f(y)|^2)dy\le2(b^2+a^2)\\ &\le2(a+b)^2=2\|f\|_{H^1}^2, \end{split} \end{equation} whence \begin{equation} |f(0)|\le\sqrt2\,\|f\|_{H^1}, \end{equation} as desired.

Using here the interval $[0,a/b]$ instead of $[0,1]$, one can improve the above inequality to \begin{equation} |f(0)|\le2\sqrt{ab}=2\sqrt{\|f\|_2\,\|f'\|_2}\le \|f\|_{H^1}. \end{equation}

Let $f:=\varphi$, $a:=\|f\|_2$, $b:=\|f'\|_2$, so that $\|f\|_{H^1}=a+b$. Without loss of generality, $x=0$. For all $y\in[0,1]$, we have $$|f(y)-f(0)|\le\int_0^y|f'(t)|dt\le\int_0^1|f'(t)|dt\le\sqrt{\int_0^1|f'(t)|^2 dt}\le b,$$ by H"older's inequality, whence $|f(0)|\le b+|f(y)|$. So, \begin{equation} f(0)^2\le\int_0^1(b+|f(y)|)^2dy\le2\int_0^1(b^2+|f(y)|^2)dy\le2(b^2+a^2)\le2(a+b)^2=2\|f\|_{H^1}^2, \end{equation} whence \begin{equation} |f(0)|\le\sqrt2\,\|f\|_{H^1}, \end{equation} as desired.

Using here the interval $[0,a/b]$ instead of $[0,1]$, one can improve the above inequality to \begin{equation} |f(0)|\le2\sqrt{ab}=2\sqrt{\|f\|_2\,\|f'\|_2}\le \|f\|_{H^1}. \end{equation}

Let $f:=\varphi$, $a:=\|f\|_2$, $b:=\|f'\|_2$, so that $\|f\|_{H^1}=a+b$; see e.g. Wikipedia for the definition of $H^k$. Without loss of generality, $x=0$. For all $y\in[0,1]$, we have $$|f(y)-f(0)|\le\int_0^y|f'(t)|dt\le\int_0^1|f'(t)|dt\le\sqrt{\int_0^1|f'(t)|^2 dt}\le b,$$ by H"older's inequality, whence $|f(0)|\le b+|f(y)|$. So, \begin{equation} f(0)^2\le\int_0^1(b+|f(y)|)^2dy\le2\int_0^1(b^2+|f(y)|^2)dy\le2(b^2+a^2)\le2(a+b)^2=2\|f\|_{H^1}^2, \end{equation} whence \begin{equation} |f(0)|\le\sqrt2\,\|f\|_{H^1}, \end{equation} as desired.

Using here the interval $[0,a/b]$ instead of $[0,1]$, one can improve the above inequality to \begin{equation} |f(0)|\le2\sqrt{ab}=2\sqrt{\|f\|_2\,\|f'\|_2}\le \|f\|_{H^1}. \end{equation}

Let $f:=\varphi$, $a:=\|f\|_2$, $b:=\|f'\|_2$, so that $\|f\|_{H^1}=a+b$. Without loss of generality, $x=0$. For all $y\in[0,1]$, we have $$|f(y)-f(0)|\le\int_0^y|f'(t)|dt\le\int_0^1|f'(t)|dt\le\sqrt{\int_0^1|f'(t)|^2 dt}\le b,$$ by H"older's inequality, whence $|f(0)|\le b+|f(y)|$. So, \begin{equation} f(0)^2\le\int_0^1(b+|f(y)|)^2dy\le2\int_0^1(b^2+|f(y)|^2)dy\le2(b^2+a^2)\le2(a+b)^2=2\|f\|_{H^1}^2, \end{equation} whence \begin{equation} |f(0)|\le\sqrt2\,\|f\|_{H^1}, \end{equation} as desired.

Let $f:=\varphi$, $a:=\|f\|_2$, $b:=\|f'\|_2$, so that $\|f\|_{H^1}=a+b$. Without loss of generality, $x=0$. For all $y\in[0,1]$, we have $$|f(y)-f(0)|\le\int_0^y|f'(t)|dt\le\int_0^1|f'(t)|dt\le\sqrt{\int_0^1|f'(t)|^2 dt}\le b,$$ by H"older's inequality, whence $|f(0)|\le b+|f(y)|$. So, \begin{equation} f(0)^2\le\int_0^1(b+|f(y)|)^2dy\le2\int_0^1(b^2+|f(y)|^2)dy\le2(b^2+a^2)\le2(a+b)^2=2\|f\|_{H^1}^2, \end{equation} whence \begin{equation} |f(0)|\le\sqrt2\,\|f\|_{H^1}, \end{equation} as desired.

Using here the interval $[0,a/b]$ instead of $[0,1]$, one can improve the above inequality to \begin{equation} |f(0)|\le2\sqrt{ab}=2\sqrt{\|f\|_2\,\|f'\|_2}\le \|f\|_{H^1}. \end{equation}