Throughout, let $(X,\mathfrak{M},\mu)$ be a metric space where $\mu$ is positive.

## The question

If $f$ is of $L^p(\mu)$, which means $\lVert f \rVert_p=\left(\int_X |f|^p d\mu\right)^{1/p}<\infty$, or equivalently $\int_X |f|^p d\mu<\infty$, then we may say $|f|^p$ is of $L^1(\mu)$. In other words, we have a functional

This functional does not have to be one to one due to absolute value. But we hope this functional to be ‘fine’ enough, at the very least, we hope it is continuous.

Here, $f \sim g$ means that $f-g$ equals to $0$ almost everywhere with respect to $\mu$.

## Continuity

We still use $\varepsilon-\delta$ argument but it’s in a metric space. Suppose $(X,d_1)$ and $(Y,d_2)$ are two metric spaces and $f:X \to Y$ is a function. We say $f$ is continuous at $x_0 \in X$ if for any $\varepsilon>0$, there exists some $\delta>0$ such that $d_2(f(x_0),f(x))<\varepsilon$ whenever $d_1(x_0,x)<\delta$. Further, we say $f$ is continuous on $X$ if $f$ is continuous at every point $x \in X$.

## Metrics

For $1\leq p<\infty$, we already have a metric by

given that $d(f,g)=0$ if and only if $f \sim g$. This is complete and makes $L^p$ a Banach space. But for $0<p<1$ (yes we are going to cover that), things are much more different, and there is one reason: Minkowski inequality holds reversely! In fact we have

for $0<p<1$. In fact, $L^p$ space has too many weird things when $0<p<1$. Precisely,

For $0<p<1$, $L^p(\mu)$ is locally convex if and only if $\mu$ assumes finitely many values. (Proof.)

On the other hand, for example, $X=[0,1]$ and $\mu=m$ be the Lebesgue measure, then $L^p(\mu)$ has *no* open convex subset other than $\varnothing$ and $L^p(\mu)$ itself. However,

A topological vector space $X$ is normable if and only if its origin has a convex bounded neighborhood. (See Kolmogorov’s normability criterion.)

Therefore $L^p(m)$ is not normable, hence not Banach.

We have gone too far. We need a metric that is fine enough.

### Metric of $L^p$ when $0<p<1$

*In this subsection we always have $0<p<1$.*

Define

for $f \in L^p(\mu)$. We will show that we have a metric by

Fix $y\geq 0$, consider the function

We have $f(0)=y^p$ and

when $x > 0$ and hence $f(x)$ is nonincreasing on $[0,\infty)$, which implies that

Hence for any $f$, $g \in L^p$, we have

This inequality ensures that

is a metric. It’s immediate that $d(f,g)=d(g,f) \geq 0$ for all $f$, $g \in L^p(\mu)$. For the triangle inequality, note that

This is translate-invariant as well since

The completeness can be verified in the same way as the case when $p>1$. In fact, this metric makes $L^p$ a locally bounded F-space.

## The continuity of $\lambda$

The metric of $L^1$ is defined by

We need to find a relation between $d_p(f,g)$ and $d_1(\lambda(f),\lambda(g))$, where $d_p$ is the metric of the corresponding $L^p$ space.

### $0<p<1$

As we have proved,

Without loss of generality we assume $x \geq y$ and therefore

Hence

By interchanging $x$ and $y$, we get

Replacing $x$ and $y$ with $|f|$ and $|g|$ where $f$, $g \in L^p$, we get

But

and we therefore have

Hence $\lambda$ is continuous (and in fact, Lipschitz continuous and uniformly continuous) when $0<p<1$.

## $1 \leq p < \infty$

It’s natural to think about Minkowski’s inequality and Hölder’s inequality in this case since they are critical inequality enablers. You need to think about some examples of how to create the condition to use them and get a fine result. In this section we need to prove that

This inequality is surprisingly easy to prove however. We will use nothing but the mean value theorem. Without loss of generality we assume that $x > y \geq 0$ and define $f(t)=t^p$. Then

where $y < \zeta < x$. But since $p-1 \geq 0$, we see $\zeta^{p-1} < x^{p-1} <x^{p-1}+y^{p-1}$. Therefore

For $x=y$ the equality holds.

Therefore

By *Hölder’s inequality*, we have

By *Minkowski’s inequality*, we have

Now things are clear. Since $1/p+1/q=1$, or equivalently $1/q=(p-1)/p$, suppose $\lVert f \rVert_p$, $\lVert g \rVert_p \leq R$, then $(p-1)q=p$ and therefore

Summing the inequalities above, we get

hence $\lambda$ is continuous.

## Conclusion and further

We have proved that $\lambda$ is continuous, and when $0<p<1$, we have seen that $\lambda$ is Lipschitz continuous. It’s natural to think about its differentiability afterwards, but the absolute value function is not even differentiable so we may have no chance. But this is still a fine enough result. For example we have no restriction to $(X,\mathfrak{M},\mu)$ other than the positivity of $\mu$. Therefore we may take $\mathbb{R}^n$ as the Lebesgue measure space here, or we can take something else.

It’s also interesting how we use elementary Calculus to solve some much more abstract problems.