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Let $1<p<\infty$ and $K$ be a closed subspace of $L^p(X, \mathcal{M}, \mu)$. If $f\in L^p$ then there exists a unique $h\in K$ such that $||f-h||_p$ equals $$ \text{dist}(f,K)=\inf_{g\in K}||f-g||_p. $$ I've seen a proof using uniform convexity via Hanner's inequality. But is there a way to show this without using Hanner's inequality?

I know that since $K$ is a closed subspace of the reflexive $L^p$, $K$ is also reflexive. Also, any bounded, closed and convex subset of $K$ or $L^p$ is compact in the weak topology (using Alaoglu's theorem).

According to the second answer here (by @ncmathsadist), "the existence of the minimum owes to a compactness result known as the Alaoglu theorem."

Also, the existence of a minimizer is Exercise 43 in Royden's Real Analysis, 4th edition, p. 180 and attributed to Beppo Levi. The exercise is in the section titled "Mimimization of Convex Functionals."

It'd be great if you could help me show this without Hanner's inequality or point me to a reference. Thank you.

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theLowerCrust
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Let $(g_{n})$ be a sequence in $K$ satisfying $\left\|f-g_{n}\right\|_{p}\rightarrow\delta:=\inf_{g\in K}\left\|f-g\right\|_{p}$. It follows from the triangle inequality that the $g_{n}$ are bounded in norm, and therefore lie in a closed, convex subset $E\subset K$, which is weakly compact by Alaoglu's theorem. Whence $(g_{n})$ has a subset $(g_{\alpha})_{\Lambda}$ which weakly converges to some $g_{*}\in E$. For any bounded linear functional $\varphi$ with $\left\|\varphi\right\|=1$, we have that

$$\left|\varphi(f-g_{*})\right|\leq\left|\varphi(f-g_{\alpha})\right|+\left|\varphi(g_{\alpha}-g_{*})\right|\leq\left\|f-g_{\alpha}\right\|_{p}+\left|\varphi(g_{\alpha}-g_{*})\right|$$

For $\epsilon>0$ given, let $\varphi$ be a norm one linear functional satisfying $\left|\varphi(f-g_{*})\right|+\epsilon>\left\|f-g_{*}\right\|_{p}$. Then

$$\left\|f-g_{*}\right\|_{p}-\epsilon<\lim\left\|f-g_{\alpha}\right\|_{p}+\lim\left|\varphi(g_{\alpha}-g_{*})\right|=\delta$$

Since $\epsilon>0$ was arbitrary, we conclude that $\left\|f-g_{*}\right\|_{p}\leq\delta$. The reverse inequality is tautological.

Matt Rosenzweig
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  • Thank you! Is the existence of the norm one linear functional satisfying $\left|\varphi(f-g_{})\right|+\epsilon>\left|f-g_{}\right|{p}$ a consequence of the Hahn-Banach theorem? Could we have obtained a norm-one linear functional satisfying $\left|\varphi(f-g{})\right|=\left|f-g_{}\right|_{p}$? – theLowerCrust Apr 01 '15 at 22:13
  • @theLowerCrust: Yes, I was implicitly using the Hahn-Banach theorem. You are right that the $\epsilon$ was unnecessary (I was invoking the identity $\left|f-g_{}\right|{p}=\sup{\left|\varphi\right|=1}\left|\varphi(f-g_{})\right|$). It is a consequence of the Hahn-Banach theorem that given any element $x$ in a normed space $X$, there exists a $\varphi\in X^{}$ such that $\left|\varphi\right|_{X^{}}=1$ and $\varphi(x)=\left|x\right|_{X}$. – Matt Rosenzweig Apr 01 '15 at 22:22
  • Could you please explain why $\left|\varphi(f-g_{\alpha})\right| \leq\left|f-g_{\alpha}\right|_{p}$ for any bounded linear $\varphi$? Thank you! (I don't think this affects your argument, but should it be fixed?) – theLowerCrust Apr 01 '15 at 23:09
  • @theLowerCrust: I forgot to specify that $\left|\varphi\right|=1$. I edited the answer. – Matt Rosenzweig Apr 01 '15 at 23:26