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I'd like to show that \begin{align} ||f||_{L^p(0,1)} \leq \lambda_p ||f'||_{L^2(0,1)} \end{align} for all $f \in H^{1}_{0}(0,1)$.

I thought about doing something like \begin{align} \Big(\int_0^1|f(t)|^pdt \Big)^{1/p} = \Big( \int_0^1 |\int_0^t f'(s)ds|^p dt \Big)^{1/p} \leq \Big( \int_0^1 \big( \int_0^t |f'(s)|ds \big)^pdt \Big)^{1/p} \end{align} but I'm not so sure if that's the way to go as I'm not sure how to proceed now. Do I need to use the Poincare inequaltiy or Sobolev embedding?

sunflower234
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1 Answers1

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I am writing @Arctic Char's comment in detail.

For $f \in H^1_0(0,1)$ we write $$|f(x)| = | \int_0^x f'(t) dt| \leq \int_0^1 |f'(t)| dt \leq \left(\int_0^1 |f'(t)|^2 dt \right)^{\frac{1}{2}} \cdot \left(\int_0^1 1^2 dt \right)^{\frac{1}{2}} = ||f'||_{L^2}. $$ Thus you get $$||f||_{\infty} \leq ||f'||_{L^2}.$$ On the other hand, we have $$||f||^p_{L^p} = \int_0^1 |f(x)|^p dx \leq \int_0^1 ||f||_{\infty}^p dx $$ Which gives you $$||f||_{L^p} \leq ||f||_{\infty}.$$

Tarek Acila
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  • Thank you! I did the same thing but I was confused since I don't seem to use the boundary conditions $f(0)=f(1)=0$ amywhere. So is this true for general $f \in H^1$? – sunflower234 Mar 19 '23 at 14:28
  • The boundary condition is used in the very first equality because we have $|\int^x_0 f'(t) dt| = |f(x) - f(0)| = |f(x)|$. – Tarek Acila Mar 19 '23 at 14:31
  • Oh yes of course! thanks – sunflower234 Mar 19 '23 at 14:41
  • I've just came across the statement that this equality is also true for $f \in H^1$ with the additional condition that $\int_0^1f(x)dx=0$. Do you know if this can be proven similarly? (I tried it but I failed) – sunflower234 Mar 19 '23 at 20:12
  • It is indeed true under the assumptions you mentioned. That is because this inequality does not work for constant functions other than $0$. In $H^1_0$ the only constant function is the $0$ function (because of the boundary condition). And under the assumption you gave, the only constant function in $H^1$ is still the $0$ function because the only constant function with average $0$ is the $0$. As per the proof, check this post. – Tarek Acila Mar 19 '23 at 21:19