Prove that if $f\in L^{p}(E), 1\leq p<\infty, m(E)<\infty$, then $f\in L^{q}(E)$ for all $1\leq q\leq p$

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Problem: Let $f\in L^{p}(\mathbb{R}^d),$ for $1\leq p<\infty,$ and $f$ is supported on a set $E$, of finite measure. Prove that $f\in L^{q}(\mathbb{R}^d)$ for all $1\leq q\leq p.$

Here are my thoughts so far:

If $p=1$, then there is nothing to prove, so suppose $p>1$. By hypothesis $$\|f\|_{p}=\left(\int_{\mathbb{R}^d}\vert f\vert^{p}\right)^{1/p}=\left(\int_{E}\vert f\vert^{p}\right)^{1/p}\leq M,$$ for some $0<M<\infty.$ Therefore, $$\int_{E}\vert f\vert^{p}\leq M^{p},$$ and hence $\vert f\vert^{p} \leq C$ for a.e. $x\in E,$ for some $0<C<\infty.$ Choosing $C$ large enough, we can conclude that $\vert f\vert\leq C^{1/p}$ for a.e. $x\in E$. Then $\vert f\vert^{q}\leq C^{q/p}$ for a.e. $x\in E$. And it follows that $$\|f\|_{q}=\left(\int_{\mathbb{R}^d}\vert f\vert^{q}\right)^{1/q}\leq m(E)^{1/q}C^{q/p}.$$ And we have that $f\in L^{q}(\mathbb{R}^d).$


Is my reasoning correct? My main concern whether or not choosing $C$ as above is correct or not, as the function could be approaching $0$. Any feedback is much appreciated.

Thank you for your time.

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Note that: $$ \|f\|_p = \left(\int_Ef^p\right)^{1/p} = \left(\int \chi_E f^p\right)^{1/p} $$ where $\chi_E$ is the characteristic function of $E$. Now, for $1 \leq q \leq p$, consider $f^q \chi_E = (f \chi_E)^q$. Take $r$ so that $\frac{1}{p} + \frac{1}{r} = \frac{1}{q}$. This is doable because $p> q$. A version of Holders inequality tells us that for $f\in L^p, g\in L^r$: $$ \|fg\|_q \leq \|f\|_p\|g\|_r $$ Hence, letting $f = f$, and $g = \chi E$, we have: $$ \|f\|_q = \|f\chi_E\|_q \leq \|f\|_p\|\chi_E\|_r = \|f\|_p\mu(E)^{1/r} $$ And the right hand side is finite thanks to $E$ having finite measure.

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We don't need Holder here. Let $A=\{x\in E: |f(x)|\le 1\},$ $B=\{x\in E: |f(x)|>1 \}.$ Note that $|f|^q \le |f|^p$ on $B.$ We conclude

$$\int_E |f|^q = \int_A |f|^q + \int_B |f|^q$$ $$ \le 1\cdot m(A) + \int_B |f|^p \le m(E) + \int_E |f|^p < \infty.$$