Prove $\| \int_0^{\tau} f(s, \cdot) ds \|_p \leq \int_0^{\tau} \| f(s,\cdot) \|_p ds $

Question

For a smooth compactly supported function $f:(0,\infty)\times \mathbb R \to \mathbb R$ is it true that $$\left\| \int_0^{t} f(s, \cdot) ds \right\|_p \leq \int_0^{t} \| f(s,\cdot) \|_p ds $$ holds?

Answer 1

Here's the general statement of the theorem:

Minkowski's Inequality for Integrals:

Let $(X,\mathfrak{M},\mu), (Y,\mathfrak{N},\nu)$ be $\sigma$-finite measure spaces. Then, for any ($\mathfrak{M}\otimes\mathfrak{N}-\mathcal{B}(\overline{\Bbb{R}})$) measurable function $f:X\times Y \to [0,\infty]$, and any $p\in [1,\infty]$, we have \begin{align} \left\lVert\int_Xf(x,\cdot)\,d\mu(x)\right\rVert_p &\leq \int_X\lVert f(x,\cdot)\rVert_p\,d\mu(x) \end{align}

This is of course completely general, and from this one can easily deduce the special cases of when the target space is $\Bbb{C}$ (or really any other finite-dimensional vector space). Also, by taking $X,Y$ to be suitable subsets of $\Bbb{R}$ equipped with the Lebesgue measure, and requiring $f$ to be smooth with compact support, we can recover the situation you're asking about. So, it is up to you how general you want to be; just note that making any simplifications from the beginning does NOT simplify the resulting proof in any way.

Roughly speaking, if you think of integrals as sums, then this is saying the norm of a sum is always less or equal to the sum of the norms; so it generalizes the triangle inequality on $L^p$ spaces. In fact, if we take $X=\{1,2\}$ to be a set containing two elements, $\mathfrak{M}$ to be the trivial $\sigma$-algebra (namely the power set of $X=\{1,2\}$), and if we let $\mu$ be the counting measure on $X$, then this inequality says that if we define $f_1(y)=f(1,y)$ and $f_2(y)=f(2,y)$ then $\lVert f_1+f_2\rVert_p\leq \lVert f_1\rVert_p+\lVert f_2\rVert_p$, i.e this reduces to precisely to the usual triangle inequality of $\lVert \cdot\rVert_p$ on the space $L^p(Y,\mathfrak{N},\nu)$.

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The Proof:

In the case of $p=1$, we actually have equality due to Tonelli's theorem (which says for non-negative, measurable functions we can ALWAYS swap the order of integration). The case $p=\infty$ is easy enough that I shall leave it to you to justify.

Now, suppose $1<p<\infty$, let $1<q<\infty$ be the Holder conjugate exponent (defined by the equation $\frac{1}{p}+\frac{1}{q}=1$) and define $g(y):=\int_Xf(x,y)\,d\mu(x)$. Then, \begin{align} \lVert g\rVert_p^p &:= \int_Yg(y) \cdot g(y)^{p-1}\,d\nu(y)\\ &:=\int_Y\int_Xf(x,y)g(y)^{p-1}\,d\mu(x)\,d\nu(y)\\ &= \int_X \left[\int_Yf(x,\cdot )g(\cdot)^{p-1}\,d\nu\right]\,d\mu(x)\tag{Tonelli's theorem}\\ &\leq\int_X \lVert f(x,\cdot)\rVert_p\cdot \lVert g^{p-1}\rVert_q\,d\mu(x)\tag{Holder's inequality}\\ &=\lVert g\rVert_p^{p/q}\cdot \int_X \lVert f(x,\cdot)\rVert_p\,d\mu(x), \end{align} where the last line follows immediately once you write out the definition of the $q$-norm as an integral (this is where I'm using the fact that $1<p,q<\infty$) of $g^{p-1}$, and use the fact that $\frac{1}{p}+\frac{1}{q}=1$ (it's a good exercise for you to verify these things). Now, if $\lVert g\rVert_p=0$, then the inequality we're trying to prove is trivially true (because $\lVert g\rVert_p$ is the LHS so if it is $0$ then the inequality holds trivially). So, suppose it is not $0$; then we can divide both sides by it to get \begin{align} \lVert g\rVert_p^{p-\frac{p}{q}} &\leq \int_X\lVert f(x,\cdot) \rVert_p\,d\mu(x). \end{align} Note that $p-\frac{p}{q}=1$, so after plugging in the definition of $g$, we get precisely what we wanted to show: \begin{align} \left\lVert\int_Xf(x,\cdot)\,d\mu(x)\right\rVert_p &\leq \int_X\lVert f(x,\cdot)\rVert_p\,d\mu(x). \end{align} This completes the proof in every case.

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Edit in Response to Comment:

In the statement of the theorem, I assumed $f$ takes values in $[0,\infty]$, because this makes the statement of the theorem simpler. In this case, Tonelli's theorem does indeed give us equality in the case of $p=1$. In the case where $f$ takes values in a (say finite-dimensional, either real or complex... doesn't really matter) Banach space $ (V,|\cdot |)$ (for example, $\Bbb{R}$ or $\Bbb{R}^n$ or $\Bbb{C}$), we do not have equality in general for $p=1$; only an inequality; also if we want to be super formal, we would have to modify the statement of the theorem slightly:

Let $(X,\mathfrak{M},\mu), (Y,\mathfrak{N},\nu)$ be $\sigma$-finite measure spaces and let $(V,|\cdot |)$ be a (say finite-dimensional) Banach space. Then, for any $p\in [1,\infty]$, and for any ($\mathfrak{M}\otimes\mathfrak{N}-\mathcal{B}(V)$) measurable function $f:X\times Y \to V$ such that for $\nu$-a.e $y\in Y$ we have $f(\cdot, y)\in L^1(\mu)$, it follows that \begin{align} \left\lVert\int_Xf(x,\cdot)\,d\mu(x)\right\rVert_p &\leq \int_X\lVert f(x,\cdot)\rVert_p\,d\mu(x) \end{align}

By the way, note that I had to add in the extra condition of $f(\cdot, y)\in L^1(\mu)$ for $\nu$-a.e $y\in Y$, because otherwise the LHS of the inequality doesn't even make any sense... this is just one of those finiteness technical details which one has to keep track of so that the statement makes sense (we don't have this issue in the non-negative case because we can just declare an integral to be $\infty$, and the theorem means if the LHS is $\infty$, so is the RHS).

Before proving this, let us for the sake of completeness record the following lemma:

Lemma:

Given any measure space $(X,\mathfrak{M},\mu)$ and (say finite-dimensional) Banach space $(V,|\cdot|)$, if $f:X\to V$ is in $L^1(\mu)$ then \begin{align} \left|\int_Xf\,d\mu\right| & \leq \int_X |f|\,d\mu. \end{align}

This is immediate for simple functions; then simply approximate any $L^1$ function using simple functions (in the case of $V=\Bbb{R}$ or $\Bbb{C}$, alternate proofs are possible, but in any case I hope you find this simple enough).

Finally, based on this lemma and my first version of the theorem for non-negative functions, the vector valued case can easily be deduced: \begin{align} \left\lVert\int_Xf(x,\cdot)\,d\mu(x)\right\rVert_p &\leq \left\lVert \int_X\left|f(x,\cdot)\right|\,d\mu(x)\right\rVert_p \leq \int_X\lVert f(x,\cdot) \rVert_p\,d\mu(x), \end{align} where the first inequality is due to the lemma, and the second is due to the non-negative version of the theorem (i.e I'm applying the theorem to the non-negative function $|f|$).